Best Approximation

Abstract

In Chap. 7 we study best approximation problems in a general Banach space. It is well known that best approximation problems have solutions only under certain assumptions on the space X. In view of the Lau-Konjagin result these assumptions cannot be removed. On the other hand, many generic results in nonlinear functional analysis hold in any Banach space. Therefore the following natural question arises: can generic results for best approximation problems be obtained in general Banach spaces? In this chapter we answer this question in the affirmative. To this end, we consider a new framework. The main feature of this new framework is that a best approximation problem is determined by a pair consisting of a point and a closed (convex) subset of a Banach space. We consider the complete metric space of such pairs equipped with a natural complete metric and show that for most (in the sense of Baire category) pairs the corresponding best approximation problem has a unique solution. We also provide some generalizations and extensions of this result.

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7.1 Well-Posedness and Porosity

Given a nonempty closed subset A of a Banach space (X,∥⋅∥) and a point x∈X, we consider the minimization problem

$$ \min\bigl\{ \|x-y\|: y \in A\bigr\} . $$

(P)

It is well known that if A is convex and X is reflexive, then problem (P) always has at least one solution. This solution is unique when X is strictly convex.

If A is merely closed but X is uniformly convex, then according to classical results of Stechkin [173] and Edelstein [59], the set of all points in X having a unique nearest point in A is G _δ and dense in X. Since then there has been a lot of activity in this direction. In particular, it is known [84, 88] that the following properties are equivalent for any Banach space X:

• (A) X is reflexive and has a Kadec-Klee norm.

• (B) For each nonempty closed subset A of X, the set of points in X∖A with nearest points in A is dense in X∖A.

• (C) For each nonempty closed subset A of X, the set of points in X∖A with nearest points in A is generic (that is, a dense G _δ subset) in X∖A.

A more recent result of De Blasi, Myjak and Papini [52] establishes well-posedness of problem (P) for a uniformly convex X, closed A and a generic x∈X.

In this connection we recall that the minimization problem (P) is said to be well posed if it has a unique solution, say a ₀, and every minimizing sequence of (P) converges to a ₀.

A more precise formulation of the De Blasi-Myjak-Papini result mentioned above involves the notion of porosity.

Using this terminology and denoting by F the set of all points such that the minimization problem (P) is well posed, we note that De Blasi, Myjak and Papini [52] proved, in fact, that the complement X∖F is σ-porous in X.

However, the fundamental restriction in all these results is that they hold only under certain assumptions on the space X. In view of the Lau-Konjagin result mentioned above these assumptions cannot be removed. On the other hand, many generic results in nonlinear functional analysis hold in any Banach space. Therefore the following natural question arises: can generic results for best approximation problems be obtained in general Banach spaces? In [138] we answer this question in the affirmative. In this chapter we present the results obtained in [138].

To this end, we change our point of view and consider a new framework. The main feature of this new framework is that the set A in problem (P) may also vary. In our first result (Theorem 7.3) we fix x and consider the space S(X) of all nonempty closed subsets of X equipped with an appropriate complete metric, say h. We then show that the collection of all sets A∈S(X) for which problem (P) is well posed has a σ-porous complement.

In the second result (Theorem 7.4) we consider the space of pairs S(X)×X with the metric h(A,B)+∥x−y∥, where A,B∈S(X) and x,y∈X. Once again we show that the family of all pairs (A,x)∈S(X)×X for which problem (P) is well-posed has a σ-porous complement.

In our third result (Theorem 7.5) we show that for any nonempty, separable and closed subset X ₀ of X, there exists a subset $\mathcal{F}$ of (S(X),h) with a σ-porous complement such that any $A \in {\mathcal{F}}$ has the following property:

There exists a dense G _δ subset F of X ₀ such that for any x∈F, the minimization problem (P) is well posed.

In order to prove these results we now provide more information on porous sets.

Let (Y,ρ) be a metric space. We denote by B _ρ (y,r) the closed ball of center y∈Y and radius r>0.

The following simple observation was made in [180].

Proposition 7.1

Let E be a subset of the metric space (Y,ρ). Assume that there exist r ₀>0 and β∈(0,1) such that the following property holds:

(P1):

For each x∈Y and each r∈(0,r ₀], there exists z∈Y∖E such that ρ(x,z)≤r and B _ρ (z,βr)∩E=∅.

Then E is porous with respect to ρ.

Proof

Let x∈Y and r∈(0,r ₀]. By property (P1), there exists z∈Y∖E such that

$$\rho(x,z) \le r/2 \quad\hbox{and}\quad B_{\rho}(z,\beta r/2) \cap E =\emptyset. $$

Hence B _ρ (z,βr/2)⊂B _ρ (x,r)∖E and Proposition 7.1 is proved. □

As a matter of fact, property (P1) can be weakened.

Proposition 7.2

Let E be a subset of the metric space (Y,ρ). Assume that there exist r ₀>0 and β∈(0,1) such that the following property holds:

(P2):

For each x∈E and each r∈(0,r ₀], there exists z∈Y∖E such that ρ(x,z)≤r and B _ρ (z,βr)∩E=∅.

Then E is porous with respect to ρ.

Proof

We may assume that β<1/2. Let x∈Y and r∈(0,r ₀]. We will show that there exists z∈Y∖E such that

$$ \rho(x,z) \le r \quad\hbox{and}\quad B_{\rho}(z,\beta r/2) \cap E= \emptyset. $$

(7.1)

If B _ρ (x,r/4)∩E=∅, then (7.1) holds with z=x. Assume now that B _ρ (x,r/4)∩E≠∅. Then there exists

$$ x_1 \in B_{\rho}(x,r/4)\cap E. $$

(7.2)

By property (P2), there exists z∈Y∖E such that

$$ \rho(x_1,z) \le r/2 \quad\hbox{and}\quad B_{\rho}(z, \beta r/2) \cap E=\emptyset. $$

(7.3)

The relations (7.2) and (7.3) imply that

$$\rho(x,z) \le\rho(x,x_1)+\rho(x_1,z) \le 3r/4. $$

Thus there indeed exists z∈Y∖E satisfying (7.1). Proposition 7.2 is now seen to follow from Proposition 7.1. □

The following definition was introduced in [180].

Assume that a set Y is equipped with two metrics ρ ₁ and ρ ₂ such that ρ ₁(x,y)≤ρ ₂(x,y) for all x,y∈Y and that the metric spaces (Y,ρ ₁) and (Y,ρ ₂) are complete.

We say that a set E⊂Y is porous with respect to the pair (ρ ₁,ρ ₂) if there exist r ₀>0 and α∈(0,1) such that for each x∈E and each r∈(0,r ₀], there exists z∈Y∖E such that ρ ₂(z,x)≤r and $B_{\rho_{1}}(z,\alpha r ) \cap E=\emptyset$.

Proposition 7.2 implies that if E is porous with respect to (ρ ₁,ρ ₂), then it is porous with respect to both ρ ₁ and ρ ₂.

A set E⊂Y is called σ-porous with respect to (ρ ₁,ρ ₂) if it is a countable union of sets which are porous with respect to (ρ ₁,ρ ₂).

As a matter of fact, it turns out that our results are true not only for Banach spaces, but also for all complete hyperbolic spaces.

Let (X,ρ,M) be a complete hyperbolic space. For each x∈X and each A⊂X, set

$$\rho(x,A)=\inf\bigl\{ \rho(x,y): y \in A\bigr\} . $$

Denote by S(X) the family of all nonempty closed subsets of X. For each A,B∈S(X), define

$$ H(A,B):=\max\bigl\{ \sup\bigl\{ \rho(x,B): x \in A\bigr\} , \sup\bigl\{ \rho(y,A): y \in B\bigr\} \bigr\} $$

(7.4)

and

$$\tilde{H}(A,B):=H(A,B) \bigl(1+H(A,B)\bigr)^{-1}. $$

It is easy to see that $\tilde{H}$ is a metric on S(X) and that the space $(S(X),\tilde{H})$ is complete.

Fix θ∈X. For each natural number n and each A,B∈S(X), we set

$$ h_n(A,B)=\sup\bigl\{ \bigl|\rho(x,A)-\rho(x,B)\bigr|: x \in X \hbox{ and } \rho(x,\theta) \le n\bigr\} $$

(7.5)

and

$$h(A,B)=\sum_{n=1}^{\infty}\bigl[2^{-n}h_n(A,B) \bigl(1+h_n(A,B)\bigr)^{-1}\bigr]. $$

Once again it is not difficult to see that h is a metric on S(X) and that the metric space (S(X),h) is complete. Clearly,

$$\tilde{H}(A,B) \ge h(A,B)\quad \hbox{for all } A,B \in S(X). $$

We equip the set S(X) with the pair of metrics $\tilde{H}$ and h.

We now state the following three results which were obtained in [138]. Their proofs are given later in this chapter.

Theorem 7.3

Let (X,ρ,M) be a complete hyperbolic space and let $\tilde{x} \in X$. Then there exists a set Ω⊂S(X) such that its complement S(X)∖Ω is σ-porous with respect to the pair $(h,\tilde{H})$ and such that for each A∈Ω, the following property holds:

(C1):

There exists a unique $\tilde{y} \in A$ such that $\rho(\tilde{x},\tilde{y})= \rho(\tilde{x},A)$. Moreover, for each ε>0, there exists δ>0 such that if x∈A satisfies $\rho(\tilde{x},x) \le\rho(\tilde{x},A)+\delta$, then $\rho(x,\tilde{y}) \le \varepsilon$.

To state the following result we endow the Cartesian product S(X)×X with the pair of metrics d ₁ and d ₂ defined by

$$\begin{aligned} &d_1\bigl((A,x),(B,y)\bigr)=h(A,B)+\rho(x,y),\\ & d_2 \bigl((A,x),(B,y)\bigr)=\tilde{H}(A,B)+\rho(x,y), \quad x,y \in X, A,B \in S(X). \end{aligned}$$

Theorem 7.4

Let (X,ρ,M) be a complete hyperbolic space. There exists a set Ω⊂S(X)×X such that its complement [S(X)×X]∖Ω is σ-porous with respect to the pair (d ₁,d ₂) and such that for each $(A, \tilde{x}) \in \varOmega$, the following property holds:

(C2):

There exists a unique $\tilde{y} \in A$ such that $\rho(\tilde{x},\tilde{y})= \rho(\tilde{x},A)$. Moreover, for each ε>0, there exists δ>0 such that if z∈X satisfies $\rho(\tilde{x},z) \le\delta$, B∈S(X) satisfies h(A,B)≤δ, and y∈B satisfies ρ(y,z)≤ρ(z,B)+δ, then $\rho(y,\tilde{y}) \le \varepsilon$.

In classical generic results the set A was fixed and x varied in a dense G _δ subset of X. In our first two results the set A is also variable. However, in our third result we show that if X ₀ is a nonempty, separable and closed subset of X, then for every fixed A in a dense G _δ subset of S(X) with a σ-porous complement, the set of all x∈X ₀ for which problem (P) is well posed contains a dense G _δ subset of X ₀.

Theorem 7.5

Let (X,ρ,M) be a complete hyperbolic space. Assume that X ₀ is a nonempty, separable and closed subset of X. Then there exists a set ${\mathcal{F}} \subset S(X)$ such that $S(X) \setminus{ \mathcal{F}}$ is σ-porous with respect to the pair $(h,\tilde{H})$ and such that for each $A \in{ \mathcal{F}}$, the following property holds:

(C3):

There exists a set F⊂X ₀ which is a countable intersection of open and everywhere dense subsets of X ₀ with the relative topology such that for each $\tilde{x} \in F$, there exists a unique $\tilde{y} \in A$ for which $\rho(\tilde{x},\tilde{y}) =\rho (\tilde{x},A)$. Moreover, if $\{y_{i}\}_{i=1}^{\infty} \subset A$ satisfies $\lim_{i \to\infty} \rho(\tilde{x},y_{i})=\rho(\tilde{x},A)$, then $y_{i} \to\tilde{y}$ as i→∞.

7.2 Auxiliary Results

Let (X,ρ,M) be a complete hyperbolic space and let S(X) be the family of all nonempty closed subsets of X.

Lemma 7.6

Let A∈S(X), $\tilde{x} \in X$ and let r,ε∈(0,1). Then there exists $\bar{x} \in X$ such that $\rho(\bar{x},A) \le r$ and for the set $\tilde{A}=A\cup\{\bar{x}\}$ the following properties hold:

$$\begin{gathered} \rho(\tilde{x},\bar{x})=\rho(\tilde{x}, \tilde{A}); \\ \textit{if}\ x \in\tilde{A}\ \textit{and}\ \rho(\tilde{x},x) \le \rho(\tilde{x},\tilde{A})+\varepsilon r/4,\ \textit{then}\ \rho(\bar{x},x)\le \varepsilon . \end{gathered} $$

Proof

If $\rho(\tilde{x},A) \le r$, then the lemma holds with $\bar{x} =\tilde{x}$ and $\tilde{A}=A\cup\{\tilde{x}\}$. Therefore we may restrict ourselves to the case where

$$ \rho(\tilde{x},A)>r. $$

(7.6)

Choose x ₀∈A such that

$$ \rho(\tilde{x},x_0) \le\rho(\tilde{x}, A)+r/2. $$

(7.7)

There exists

$$ \bar{x} \in\bigl\{ \gamma\tilde{x}\oplus(1-\gamma)x_0: \gamma\in(0,1)\bigr\} $$

(7.8)

such that

$$ \rho(\bar{x},x_0)=r \quad\hbox{and}\quad \rho(\tilde{x},\bar{x})= \rho(\tilde{x},x_0)-r. $$

(7.9)

Set $\tilde{A}=A\cup\{\bar{x}\}$. We have by (7.9) and (7.7),

$$\rho(\tilde{x},\bar{x}) =\rho(\tilde{x},x_0)-r \le \rho(\tilde{x},A)+r/2-r=\rho(\tilde{x},A)-r/2. $$

Therefore $\rho(\tilde{x},\bar{x})=\rho(\tilde{x}, \tilde{A})$, and if $x \in\tilde{A}$ and $\rho(\tilde{x},x)<\rho(\tilde{x}, \tilde{A})+r/2$, then $x=\bar{x}$. This completes the proof of Lemma 7.6. □

Before stating our next lemma we choose, for each ε∈(0,1) and each natural number n, a number

$$ \alpha(\varepsilon ,n) \in\bigl(0,16^{-n-2}\varepsilon \bigr). $$

(7.10)

Lemma 7.7

Let A∈S(X), $\tilde{x} \in X$ and let r,ε∈(0,1). Suppose that n is a natural number, let

$$ \alpha=\alpha(\varepsilon ,n) $$

(7.11)

and assume that

$$ \rho(\tilde{x}, \theta) \le n \quad\textit{and}\quad \bigl\{ x \in X: \rho(x, \theta) \le n\bigr\} \cap A \neq\emptyset. $$

(7.12)

Then there exists $\bar{x} \in X$ such that $\rho(\bar{x},A) \le r$ and such that the set $\tilde{A}=A\cup\{ \bar{x}\}$ has the following two properties:

$$ \rho(\tilde{x},\bar{x})=\rho(\tilde{x}, \tilde{A}); $$

(7.13)

if

$$\begin{gathered} \tilde{y} \in X,\quad \rho(\tilde{y},\tilde{x}) \le\alpha r, \end{gathered}$$

(7.14)

$$\begin{gathered} B \in S(X),\quad h(\tilde{A},B) \le\alpha r, \end{gathered}$$

(7.15)

and

$$ z \in B,\quad \rho(\tilde{y}, z) \le\rho(\tilde{y}, B)+\varepsilon r/16, $$

(7.16)

then

$$ \rho(z,\bar{x}) \le \varepsilon . $$

(7.17)

Proof

By Lemma 7.6, there exists $\bar{x} \in X$ such that

$$ \rho(\bar{x},A) \le r $$

(7.18)

and such that for the set $\tilde{A}=A\cup\{\bar{x}\}$, equality (7.13) is true and the following property holds:

$$ \hbox{If } x \in\tilde{A} \hbox{ and } \rho(\tilde{x},x) \le\rho ( \tilde{x}, \tilde{A})+\varepsilon r /8, \hbox{ then } \rho(\bar{x},x) \le \varepsilon /2. $$

(7.19)

Assume that $\tilde{y} \in X$ satisfies (7.14) and B∈S(X) satisfies (7.15). We will show that

$$ \rho(\tilde{y},B) <\rho(\tilde{x}, \tilde{A})+4\alpha r 16^n. $$

(7.20)

By (7.14),

$$\bigl|\rho(\tilde{y},\tilde{A})-\rho(\tilde{x},\tilde{A})\bigr| \le\alpha r. $$

When combined with (7.13), this implies that

$$ \bigl|\rho(\tilde{y},\tilde{A})-\rho(\tilde{x},\bar{x})\bigr| \le\alpha r. $$

(7.21)

Relations (7.13) and (7.12) imply that

$$ \rho(\tilde{x},\bar{x}) \le\rho(\tilde{x},A) \le2n \quad\hbox{and}\quad \rho( \bar{x} ,\theta) \le3n. $$

(7.22)

It follows from (7.5) and (7.15) that

$$h_{4n}(\tilde{A},B) \bigl(1+h_{4n}(\tilde{A},B) \bigr)^{-1} \le2^{4n}h(\tilde{A},B) \le2^{4n} \alpha r. $$

When combined with (7.10) and (7.11), this inequality implies that

$$ h_{4n}(\tilde{A},B) \le2^{4n}\alpha r \bigl(1-2^{4n}\alpha r\bigr)^{-1} < 2^{4n+1} \alpha r. $$

(7.23)

Since $\bar{x} \in\tilde{A}$, it now follows from (7.23), (7.22) and (7.5) that $\rho(\bar{x},B)<2^{4n+1}\alpha r$ and there exists $\bar{y} \in X$ such that

$$ \bar{y} \in B \quad\hbox{and}\quad \rho(\bar{x}, \bar{y}) <2\alpha r 16^n. $$

(7.24)

By (7.24), (7.14) and (7.13),

$$\begin{aligned} \rho(\tilde{y},B) \le&\rho(\tilde{y},\bar{y}) \le\rho(\tilde{y},\bar{x})+ \rho(\bar{x}, \bar{y}) \\ <&\rho(\tilde{y},\tilde{x})+\rho(\tilde{x},\bar{x})+2\alpha r 16^n \\ \le&2\alpha r 16^{n}+\alpha r+\rho(\tilde{x}, \tilde{A}). \end{aligned}$$

This certainly implies (7.20), as claimed.

Assume now that z∈B satisfies (7.16). It follows from (7.16), (7.20), (7.11) and (7.10) that

$$\begin{aligned} \rho(\tilde{y},z) \le&\rho(\tilde{y},B)+ \varepsilon r /16 \le \rho( \tilde{x},\tilde{A})+4\alpha r 16^n +\varepsilon r /16 \\ \le& \rho(\tilde{x}, \tilde{A})+\varepsilon r /8. \end{aligned}$$

(7.25)

Relations (7.25), (7.22) and (7.14) imply that

$$ \rho(\tilde{y},z) \le \rho(\tilde{x},\tilde{A})+\varepsilon r /8 \le2n +r/8. $$

(7.26)

By (7.26), (7.14), (7.11) and (7.12),

$$\begin{aligned} \rho(z,\theta) \le&\rho(z,\tilde{y})+\rho(\tilde{y}, \theta) \le 2n+r/8+\rho(\tilde{y},\theta) \\ \le&2n+r/8+\rho(\tilde{y},\tilde{x})+\rho(\tilde{x},\theta) \\ \le&2n +r/8+\alpha r +n \le4n. \end{aligned}$$

(7.27)

It follows from (7.23), (7.5), (7.16) and (7.27) that

$$\rho(z,\tilde{A}) =\bigl|\rho(z,\tilde{A})-\rho(z,B)\bigr| \le h_{4n}(\tilde{A},B) < 2\alpha r 16^n. $$

Hence there exists $\tilde{z} \in X$ such that

$$ \tilde{z} \in\tilde{A} \quad\hbox{and} \quad\rho(z,\tilde{z})<2\alpha r 16^n. $$

(7.28)

By (7.14), (7.28) and (7.16) we have

$$\begin{aligned} \rho(\tilde{x},\tilde{z}) \le&\rho(\tilde{x},\tilde{y})+ \rho(\tilde{y},z)+ \rho(z, \tilde{z}) \\ \le&\alpha r+\rho(\tilde{y},z)+2 \alpha r 16^n \\ \le&\alpha r +2\alpha r 16^n+\rho(\tilde{y}, B)+\varepsilon r/16. \end{aligned}$$

It follows from this inequality, (7.20), (7.11) and (7.10) that

$$\begin{aligned} \rho(\tilde{x},\tilde{z}) \le& \alpha r +2\alpha r 16^n+\varepsilon r/16+ \rho(\tilde{x}, \tilde{A})+4\alpha r 16^n \\ \le&\rho(\tilde{x}, \tilde{A})+8 \alpha r 16^n+\varepsilon r/16 \le\rho( \tilde{x}, \tilde{A})+\varepsilon r/8. \end{aligned}$$

Thus

$$\rho(\tilde{x},\tilde{z}) \le \rho(\tilde{x}, \tilde{A})+\varepsilon r/8. $$

Using this inequality, (7.28) and (7.19), we see that $\rho(\bar{x},\tilde{z}) \le \varepsilon /2$. Combining this fact with (7.28), (7.11) and (7.10), we conclude that

$$\rho(z,\bar{x}) \le\rho(z,\tilde{z})+\rho(\tilde{z}, \bar{x}) \le 2 \alpha r 16^n +\varepsilon /2 \le \varepsilon . $$

Thus (7.17) holds and Lemma 7.7 is proved. □

7.3 Proofs of Theorems 7.3–7.5

Proof of Theorem 7.3

For each integer k≥1, denote by Ω _k the set of all A∈S(X) which have the following property:

(P3):

There exist x _A ∈X and δ _A >0 such that if x∈A satisfies $\rho(x,\tilde{x}) \le\rho(\tilde{x},A)+\delta_{A}$, then ρ(x,x _A )≤1/k.

Clearly, Ω _k+1⊂Ω _k , k=1,2,…. Set

$$\varOmega =\bigcap_{k=1}^{\infty} \varOmega _k. $$

First we will show that S(X)∖Ω is σ-porous with respect to the pair $(h,\tilde{H})$. To meet this goal it is sufficient to show that S(X)∖Ω _k is σ-porous with respect to $(h,\tilde{H})$ for all sufficiently large integers k.

There exists a natural number k ₀ such that $\rho(\theta,\tilde{x}) \le k_{0}$. Let k≥k ₀ be an integer. We will show that the set S(X)∖Ω _k is σ-porous with respect to $(h,\tilde{H})$. For each integer n≥k ₀, set

$$E_{nk}=\bigl\{ A \in S(X) \setminus \varOmega _k: \bigl\{ z \in X: \rho(z,\theta) \le n\bigr\} \cap A \neq\emptyset\bigr\} . $$

By Lemma 7.7, the set E _nk is porous with respect to $(h,\tilde{H})$ for all integers n≥k ₀. Since $S(X) \setminus \varOmega _{k}=\bigcup_{n=k_{0}}^{\infty} E_{nk}$, we conclude that S(X)∖Ω _k is σ-porous with respect to $(h,\tilde{H})$. Therefore S(X)∖Ω is also σ-porous with respect to $(h,\tilde{H})$.

Let A∈Ω be given. We will show that A has property (C1). By the definition of Ω _k and property (P3), for each integer k≥1, there exist x _k ∈X and δ _k >0 such that the following property holds:

(P4):

If x∈A satisfies $\rho(x,\tilde{x}) \le\rho(\tilde{x},A)+\delta_{k}$, then ρ(x,x _k )≤1/k.

Let $\{z_{i}\}_{i=1}^{\infty} \subset A$ be such that

$$ \lim_{i \to\infty} \rho(\tilde{x},z_i)=\rho( \tilde{x}, A). $$

(7.29)

Fix an integer k≥1. It follows from property (P4) that for all large enough natural numbers i,

$$\rho(\tilde{x},z_i) \le\rho(\tilde{x},A)+\delta_k $$

and

$$\rho(z_i,x_k) \le1/k. $$

Since k is an arbitrary natural number, we conclude that $\{z_{i}\}_{i=1}^{\infty}$ is a Cauchy sequence which converges to some $\tilde{y} \in A$. It is clear that $\rho(\tilde{x}, \tilde{y})=\rho(\tilde{x}, A)$. If the minimizer $\tilde{y}$ were not unique, we would be able to construct a nonconvergent minimizing sequence $\{z_{i}\}_{i=1}^{\infty}$. Thus $\tilde{y}$ is the unique solution to problem (P) (with $x= \tilde{x}$) and any sequence $\{z_{i}\}_{i=1}^{\infty} \subset A$ satisfying (7.29) converges to $\tilde{y}$. This completes the proof of Theorem 7.3. □

Proof of Theorem 7.4

For each integer k≥1, denote by Ω _k the set of all $(A,\tilde{x}) \in S(X) \times X$ which have the following property:

(P5):

There exist $\bar{x} \in X$ and $\bar{\delta}>0$ such that if x∈X satisfies $\rho(x,\tilde{x}) \le\bar{\delta}$, B∈S(X) satisfies $h(A,B) \le\bar{\delta}$, and y∈B satisfies $\rho(y,x) \le\rho(x, B)+\bar{\delta}$, then $\rho(y,\bar{x}) \le1/k$.

Clearly Ω _k+1⊂Ω _k , k=1,2,… . Set

$$\varOmega =\bigcap_{k=1}^{\infty} \varOmega _k. $$

First we will show that [S(X)×X]∖Ω is σ-porous with respect to the pair (d ₁,d ₂). For each pair of natural numbers n and k, set

$$E_{nk}=\bigl\{ (A,x) \in\bigl[S(X)\times X\bigr] \setminus \varOmega _k: \rho(x,\theta) \le n, B_{\rho}(\theta,n) \cap A \neq\emptyset\bigr\} . $$

By Lemma 7.7, the set E _nk is porous with respect to (d ₁,d ₂) for all natural numbers n and k. Since

$$\bigl[S(X) \times X\bigr]\setminus \varOmega =\bigcup_{k=1}^{\infty} \bigl(\bigl[S(X) \times X\bigr]\setminus \varOmega _k\bigr) = \bigcup_{k=1}^{\infty} \bigcup_{n=1}^{\infty} E_{nk}, $$

the set [S(X)×X]∖Ω is σ-porous with respect to (d ₁,d ₂), by definition.

Let $(A,\tilde{x}) \in \varOmega$. We will show that $(A,\tilde{x})$ has property (C2).

By the definition of Ω _k and property (P5), for each integer k≥1, there exist x _k ∈X and δ _k >0 with the following property:

(P6):

If x∈X satisfies $\rho(x,\tilde{x}) \le\delta_{k}$, B∈S(X) satisfies h(A,B)≤δ _k , and y∈B satisfies ρ(y,x)≤ρ(x,B)+δ _k , then ρ(y,x _k )≤1/k.

Let $\{z_{i}\}_{i=1}^{\infty} \subset A$ be such that

$$ \lim_{i \to\infty} \rho(\tilde{x},z_i)=\rho( \tilde{x}, A). $$

(7.30)

Fix an integer k≥1. It follows from property (P6) that for all large enough natural numbers i,

$$\rho(\tilde{x},z_i) \le\rho(\tilde{x},A)+\delta_k $$

and

$$\rho(z_i,x_k) \le1/k. $$

Since k is an arbitrary natural number, we conclude that $\{z_{i}\}_{i=1}^{\infty}$ is a Cauchy sequence which converges to some $\tilde{y} \in A$. Clearly, $\rho(\tilde{x}, \tilde{y})=\rho(\tilde{x}, A)$. It is not difficult to see that $\tilde{y}$ is the unique solution to the minimization problem (P) with $x= \tilde{x}$.

Let ε>0 be given. Choose an integer k>4/min{1,ε}. By property (P6),

$$ \rho(\tilde{y},x_k) \le1/k. $$

(7.31)

Assume that z∈X satisfies $\rho(z,\tilde{x}) \le\delta_{k}$, B∈S(X) satisfies h(A,B)≤δ _k and y∈B satisfies ρ(y,z)≤ρ(z,B)+δ _k . Then it follows from property (P6) that ρ(y,x _k )≤1/k. When combined with (7.31), this implies that $\rho(y,\tilde{y}) \le2/k <\varepsilon$. This completes the proof of Theorem 7.4. □

Proof of Theorem 7.5

Let $\{x_{i}\}_{i=1}^{\infty} \subset X_{0}$ be an everywhere dense subset of X ₀. For each natural number p, there exists a set ${\mathcal{F}}_{p} \subset S(X)$ such that Theorem 7.3 holds with $\tilde{x}=x_{p}$ and $\varOmega ={\mathcal{F}}_{p}$. Set ${\mathcal{F}}=\bigcap_{p=1}^{\infty} {\mathcal{F}}_{p}$. Clearly, $S(X) \setminus {\mathcal{F}}$ is σ-porous with respect to the pair $(h,\tilde{H})$.

Let $A \in{ \mathcal{F}}$ and let p≥1 be an integer. By Theorem 7.3, which holds with $\tilde{x}=x_{p}$ and $\varOmega ={\mathcal{F}}_{p}$, there exists a unique $\bar{x}_{p} \in A$ such that

$$ \rho(x_p,\bar{x}_p)=\rho(x_p,A) $$

(7.32)

and the following property holds:

(P7):

For each integer k≥1, there exists δ(p,k)>0 such that if x∈A satisfies ρ(x,x _p )≤ρ(x _p ,A)+4δ(p,k), then $\rho(x,\bar{x}_{p}) \le1/k$.

For each pair of natural numbers p and k, set

$$V(p,k)=\bigl\{ z \in X_0: \rho(z, x_p)<\delta(p,k)\bigr\} . $$

It follows from property (P7) that for each pair of integers p,k≥1, the following property holds:

(P8):

If x∈A, z∈X ₀, ρ(z,x _p )≤δ(p,k) and ρ(z,x)≤ρ(z,A)+δ(p,k), then $\rho(x,\bar{x}_{p}) \le1/k$.

Set

$$F:=\bigcap_{k=1}^{\infty} \Bigl[\bigcup\bigl\{ V(p,k): p=1,2,\dots\bigr\} \Bigr]. $$

Clearly, F is a countable intersection of open and everywhere dense subsets of X ₀.

Let x∈F be given. Consider a sequence $\{x_{i}\}_{i=1}^{\infty} \subset A$ such that

$$ \lim_{i \to\infty}\rho(x,x_i)=\rho(x,A). $$

(7.33)

Let ε>0. Choose a natural number k>8⁻¹/min{1,ε}. There exists an integer p≥1 such that x∈V(p,k). By the definition of V(p,k), ρ(x,x _p )<δ(p,k). It follows from this inequality and property (P8) that for all sufficiently large integers i, ρ(x,x _i )≤ρ(x,A)+δ(p,k) and $\rho(x_{i},\bar{x}_{p}) \le1/k <\varepsilon /2$. Since ε is an arbitrary positive number, we conclude that $\{x_{i}\}_{i=1}^{\infty}$ is a Cauchy sequence which converges to $\tilde{y} \in A$. Clearly, $\tilde{y}$ is the unique minimizer of the minimization problem z→ρ(x,z), z∈A. Note that we have shown that any sequence $\{x_{i}\}_{i=1}^{\infty} \subset A$ satisfying (7.33) converges to $\tilde{y}$. This completes the proof of Theorem 7.5. □

7.4 Generalized Best Approximation Problems

Given a closed subset A of a Banach space X, a point x∈X and a continuous function f:X→R ¹, we consider the problem of finding a solution to the minimization problem min{f(x−y):y∈A}. For a fixed function f, we define an appropriate complete metric space $\mathcal{M}$ of all pairs (A,x) and construct a subset Ω of $\mathcal{M}$, which is a countable intersection of open and everywhere dense sets such that for each pair in Ω, our minimization problem is well posed.

Let (X,∥⋅∥) be a Banach space and let f:X→R ¹ be a continuous function. Assume that

$$\begin{gathered} \inf\bigl\{ f(x): x \in X\bigr\} \hbox{ is attained at a unique point } x_* \in X, \end{gathered}$$

(7.34)

$$\begin{gathered} \lim_{\|u\| \to\infty}f(u)=\infty, \end{gathered}$$

(7.35)

$$\begin{gathered} \hbox{if } \{x_i\}_{i=1}^{\infty} \subset X \hbox{ and } \lim_{i \to\infty}f(x_i)=f(x_*), \hbox{ then }\lim_{i \to\infty} x_i=x_*, \end{gathered}$$

(7.36)

and that for each integer n≥1, there exists an increasing function ϕ _n :(0,1)→(0,1) such that

$$ f\bigl(\alpha x+(1-\alpha)x_*\bigr) \le\phi_n ( \alpha)f(x) +\bigl(1-\phi_n(\alpha)\bigr)f(x_*) $$

(7.37)

for all x∈X satisfying ∥x∥≤n and all α∈(0,1). It is clear that (7.37) holds if f is convex.

Given a closed subset A of X and a point x∈X, we consider the minimization problem

$$ \min\bigl\{ f(x-y): y \in A\bigr\} . $$

(P)

This problem was studied by many mathematicians mostly in the case where f(x)=∥x∥. We recall that the minimization problem (P) is said to be well posed if it has a unique solution, say a ₀, and every minimizing sequence of (P) converges to a ₀. In other words, if $\{y_{i}\}_{i=1}^{\infty} \subset A$ and lim _i→∞ f(x−y _i )=f(x−a ₀), then lim _i→∞ y _i =a ₀.

Note that in the studies of problem (P) [52, 59, 84, 88, 173], the function f is the norm of the space X. There are some additional results in the literature where either f is a Minkowski functional [51, 93] or the function ∥x−y∥, y∈A, is perturbed by some convex function [42].

However, the fundamental restriction in all these results is that they only hold under certain assumptions on either the space X or the set A. In view of the Lau-Konjagin result mentioned above, these assumptions cannot be removed. On the other hand, many generic results in nonlinear functional analysis hold in any Banach space. Therefore a natural question is whether generic existence results for best approximation problems can be obtained for general Banach spaces. Positive answers to this question in the special case where f=∥⋅∥ can be found in Sects. 7.1–7.3. In the next sections, which are based on [143], we answer this question in the affirmative for a general function f satisfying (7.34)–(7.37).

To this end, we change our point of view and consider another framework, the main feature of which is that the set A in problem (P) can also vary. We prove four theorems which were established in [143]. In our first result (Theorem 7.8), we fix x and consider the space S(X) of all nonempty closed subsets of X equipped with an appropriate complete metric, say h. We then show that the collection of all sets A∈S(X) for which problem (P) is well posed contains an everywhere dense G _δ set. In the second result (Theorem 7.9), we consider the space of pairs S(X)×X with the metric h(A,B)+∥x−y∥, A,B∈S(X), x,y∈X. Once again, we show that the family of all pairs (A,x)∈S(X)×X for which problem (P) is well posed contains an everywhere dense G _δ set. In our third result (Theorem 7.10), we show that for any separable closed subset X ₀ of X, there exists an everywhere dense G _δ subset $\mathcal{F}$ of (S(X),h) such that any $A \in \mathcal{F}$ has the following property: there exists a G _δ dense subset F of X ₀ such that for any x∈F, problem (P) is well posed.

In our fourth result (Theorem 7.11), we show that a continuous coercive convex f:X→R ¹ which has a unique minimizer and a certain well-posedness property (on the whole space X) has a unique minimizer and the same well-posedness property on a generic closed subset of X.

7.5 Theorems 7.8–7.11

We recall that (X,∥⋅∥) is a Banach space, f:X→R ¹ is a continuous function satisfying (7.34)–(7.36) and that for each integer n≥1, there exists an increasing function ϕ _n :(0,1)→(0,1) such that (7.37) is true.

For each x∈X and each A⊂X, set

$$ \rho(x,A)=\inf\bigl\{ \rho(x,y): y \in A\bigr\} $$

(7.38)

and

$$ \rho_f(x,A)=\inf\bigl\{ f(x-y): y \in A\bigr\} . $$

(7.39)

Denote by S(X) the collection of all nonempty closed subsets of X. For each A,B∈S(X), define

$$ H(A,B):=\max\bigl\{ \sup\bigl\{ \rho(x,B): x \in A\bigr\} , \sup\bigl\{ \rho(y,A): y \in B\bigr\} \bigr\} $$

(7.40)

and

$$\tilde{H}(A,B):=H(A,B) \bigl(1+H(A,B)\bigr)^{-1}. $$

Here we use the convention that ∞/∞=1.

It is not difficult to see that the metric space $(S(X),\tilde{H})$ is complete.

For each natural number n and each A,B∈S(X), we set

$$ h_n(A,B):=\sup\bigl\{ \bigl|\rho(x,A)-\rho(x,B)\bigr|: x \in X \hbox{ and } \|x\| \le n\bigr\} $$

(7.41)

and

$$h(A,B):=\sum_{n=1}^{\infty}\bigl[2^{-n}h_n(A,B) \bigl(1+h_n(A,B)\bigr)^{-1}\bigr]. $$

Once again, it is not difficult to see that h is a metric on S(X) and that the metric space (S(X),h) is complete. Clearly, $\tilde{H}(A,B) \ge h(A,B)$ for all A,B∈S(X).

We equip the set S(X) with the pair of metrics $\tilde{H}$ and h. The topologies induced by the metrics $\tilde{H}$ and h on S(X) will be called the strong topology and the weak topology, respectively.

We now state Theorems 7.8–7.11.

Theorem 7.8

Let $\tilde{x} \in X$. Then there exists a set Ω⊂S(X), which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of S(X), such that for each A∈Ω, the following property holds:

(C1):

There exists a unique $\tilde{y} \in A$ such that $f(\tilde{x}- \tilde{y})= \rho_{f} (\tilde{x},A)$. Moreover, for each ε>0, there exists δ>0 such that if x∈A satisfies $f(\tilde{x}-x) \le\rho_{f}(\tilde{x},A)+\delta$, then $\|x-\tilde{y}\| \le \varepsilon$.

To state our second result we endow the Cartesian product S(X)×X with the pair of metrics d ₁ and d ₂ defined by

$$\begin{aligned} &d_1\bigl((A,x),(B,y)\bigr)=h(A,B)+\rho(x,y), \\ & d_2 \bigl((A,x),(B,y)\bigr)=\tilde{H}(A,B)+\rho(x,y), \quad x,y \in X, A,B \in S(X). \end{aligned}$$

We will refer to the topologies induced on S(X)×X by d ₂ and d ₁ as the strong and weak topologies, respectively.

Theorem 7.9

There exists a set Ω⊂S(X)×X, which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of S(X)×X, such that for each $(A, \tilde{x}) \in \varOmega$, the following property holds:

(C2):

There exists a unique $\tilde{y} \in A$ such that $f(\tilde{x}-\tilde{y})= \rho_{f}(\tilde{x},A)$. Moreover, for each ε>0, there exists δ>0 such that if z∈X satisfies $\|z-\tilde{x}\| \le\delta$, B∈S(X) satisfies h(A,B)≤δ, and y∈B satisfies f(z−y)≤ρ _f (z,B)+δ, then $\|y-\tilde{y}\| \le \varepsilon$.

In most classical generic results the set A was fixed and x varied in a dense G _δ subset of X. In our first two results the set A is also variable. However, our third result shows that for every fixed A in a dense G _δ subset of S(X), the set of all x∈X for which problem (P) is well posed contains a dense G _δ subset of X.

Theorem 7.10

Assume that X ₀ is a closed separable subset of X. Then there exists a set ${\mathcal{F}} \subset S(X)$, which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of S(X), such that for each $A \in \mathcal{F}$, the following property holds:

(C3):

There exists a set F⊂X ₀, which is a countable intersection of open and everywhere dense subsets of X ₀ with the relative topology, such that for each $\tilde{x} \in F$, there exists a unique $\tilde{y} \in A$ for which $f(\tilde{x}-\tilde{y}) =\rho_{f}(\tilde{x},A)$. Moreover, if $\{y_{i}\}_{i=1}^{\infty} \subset A$ satisfies $\lim_{i \to\infty} f(\tilde{x}-y_{i})=\rho_{f}(\tilde{x},A)$, then $y_{i} \to\tilde{y}$ as i→∞.

Now we will show that Theorem 7.8 implies the following result.

Theorem 7.11

Assume that g:X→R ¹ is a continuous convex function such that inf{g(x):x∈X} is attained at a unique point y _∗∈X, lim_{∥u∥→∞} g(u)=∞, and if $\{y_{i}\}_{i=1}^{\infty} \subset X$ and lim _i→∞ g(y _i )=g(y _∗), then y _i →y _∗ as i→∞. Then there exists a set Ω⊂S(X), which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of S(X), such that for each A∈Ω, the following property holds:

(C4):

There is a unique y _A ∈A such that g(y _A )=inf{g(y):y∈A}. Moreover, for each ε>0, there exists δ>0 such that if y∈A satisfies g(y)≤g(y _A )+δ, then ∥y−y _A ∥≤ε.

Proof

Define f(x)=g(−x), x∈X. Clearly, f is convex and satisfies (7.34)–(7.36). Therefore Theorem 7.8 is valid with $\tilde{x}=0$ and there exists a set Ω⊂S(X), which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of S(X), such that for each A∈Ω, the following property holds:

There is a unique $\tilde{y} \in A$ such that

$$g(\tilde{y})=f(-\tilde{y})=\inf\bigl\{ f(-y): y \in A\bigr\} = \inf\bigl\{ g(y): y \in A\bigr\} . $$

Moreover, for each ε>0, there exists δ>0 such that if x∈A satisfies

$$g(x)=f(-x) \le\rho_f(0,A)+\delta=\inf\bigl\{ f(-y): y \in A\bigr\} + \delta=\inf\bigl\{ g(y): y \in A\bigr\} +\delta, $$

then $\|x-\tilde{y}\| \le \varepsilon$. Theorem 7.11 is proved. □

It is easy to see that in the proofs of Theorems 7.8–7.10 we may assume without loss of generality that inf{f(x):x∈X}=0. It is also not difficult to see that we may assume without loss of generality that x _∗=0. Indeed, instead of the function f(⋅) we can consider f(⋅+x _∗). This new function also satisfies (7.34)–(7.37). Once Theorems 7.8–7.10 are proved for this new function, they will also hold for the original function f because the mapping (A,x)→(A,x+x _∗), (A,x)∈S(X)×A, is an isometry with respect to both metrics d ₁ and d ₂.

7.6 A Basic Lemma

Lemma 7.12

Let A∈S(X), $\tilde{x} \in X$, and let r,ε∈(0,1). Then there exists $\tilde{A} \in S(X)$, $\bar{x} \in\tilde{A}$, and δ>0 such that

$$ \tilde{H}(A,\tilde{A}) \le r,\quad\quad f(\tilde{x}-\bar{x})= \rho_f(\tilde{x}, \tilde{A}), $$

(7.42)

and such that the following property holds:

For each $\tilde{y} \in X$ satisfying $\|\tilde{y} -\tilde{x}\| \le \delta$, each B∈S(X) satisfying $h(B,\tilde{A}) \le\delta$, and each z∈B satisfying

$$ f(\tilde{y} -z) \le\rho_f(\tilde{y},B)+\delta, $$

(7.43)

the inequality $\|z-\bar{x}\| \le \varepsilon$ holds.

Proof

There are two cases: either $\rho(\tilde{x}, A) \le r$ or $\rho(\tilde{x}, A) > r$. Consider the first case where

$$ \rho(\tilde{x}, A) \le r. $$

(7.44)

Set

$$ \bar{x} =\tilde{x} \quad\hbox{and} \quad\tilde{A}=A\cup\{\tilde{x}\}. $$

(7.45)

Clearly, (7.42) is true. Fix an integer $n >\|\tilde{x}\|$. By (7.36), there is ξ∈(0,1) such that

$$ \hbox{if } z \in X \hbox{ and } f(z) \le4 \xi, \hbox{ then } \|z\| \le \varepsilon /2. $$

(7.46)

Using (7.34), we choose a number δ∈(0,1) such that

$$ \delta<2^{-n-4}\min\{\varepsilon ,\xi\} $$

(7.47)

and

$$ \hbox{if } z \in X \hbox{ and } \|z\| \le2^{n+4}\delta, \hbox{ then } f(z) \le\xi. $$

(7.48)

Let

$$ \tilde{y} \in X,\quad \|\tilde{y}-\tilde{x}\| \le\delta,\quad\quad B \in S(X),\quad h(B, \tilde{A}) \le\delta $$

(7.49)

and let z∈B satisfy (7.43). By (7.49) and (7.41), $h_{n}(\tilde{A},B)(1+h_{n}(\tilde{A},B))^{-1} \le2^{n} \delta$. This implies that $h_{n}(\tilde{A},B)(1-2^{n}\delta) \le2^{n}\delta$. When combined with (7.47), this inequality shows that $h_{n}(\tilde{A},B) \le2^{n+1}\delta$. Since $n >\|\tilde{x}\|$, the last inequality, when combined with (7.44) and (7.41), implies that $\rho(\tilde{x},B) \le2^{n+1}\delta$. Hence there is x ₀∈B such that $\|\tilde{x} -x_{0}\| \le2^{n+2}\delta$. This inequality and (7.49) imply in turn that $\|\tilde{y} -x_{0}\| \le2^{n+3} \delta$. The definition of δ (see (7.48)) now shows that $f(\tilde{y}-x_{0}) \le\xi$. Combining this inequality with (7.43), (7.47) and the inclusion x ₀∈B, we see that

$$ f(\tilde{y}-z) \le\delta+f(\tilde{y}-x_0) \le\xi+ \delta\le2\xi. $$

(7.50)

It now follows from (7.46) that $\|z-\tilde{y}\| \le \varepsilon /2$. Hence (7.47), (7.49) and (7.45) imply that $\|\bar{x}-z\|\le \varepsilon$. This concludes the proof of the lemma in the first case.

Now we turn our attention to the second case where

$$ \rho(\tilde{x}, A) > r. $$

(7.51)

For each t∈[0,r], set

$$ A_t=\bigl\{ v \in X: \rho(v,A) \le t\bigr\} \in S(X) $$

(7.52)

and

$$ \mu(t)=\rho_f(\tilde{x},A_t). $$

(7.53)

By (7.51) and (7.36),

$$ \mu(t)>0, \quad t \in[0,r]. $$

(7.54)

It is clear that μ(t), t∈[0,r], is a decreasing function. Choose a number

$$ t_0 \in(0,r/4) $$

(7.55)

such that μ is continuous at t ₀. By (7.35), there exists a natural number n which satisfies the following conditions:

$$ n >4\|\tilde{x}\|+8 $$

(7.56)

and

$$ \hbox{if } z \in X, f(x) \le\mu(0)+1, \hbox{ then } \|z\| \le n/4. $$

(7.57)

Let ϕ _n :(0,1)→(0,1) be an increasing function for which (7.37) is true. Choose a positive number γ∈(0,1) such that

$$ \gamma< \mu(t_0) \bigl(1-\phi(1-2r/n)\bigr)/8. $$

(7.58)

Next, choose a positive number δ ₀<1/4 such that

$$\begin{gathered} 2^{n+3}\delta_0 <\min\{\varepsilon ,\gamma\}, \end{gathered}$$

(7.59)

$$\begin{gathered}{} [t_0-4\delta_0,t_0+4 \delta_0] \subset(0,r/4), \end{gathered}$$

(7.60)

and

$$ \bigl|\mu(t)-\mu(t_0)\bigr|\le\gamma, \quad t \in[t_0-4 \delta_0,t_0+4\delta_0]. $$

(7.61)

Finally, choose a vector x ₀ such that

$$ x_0 \in A_{t_0}\quad \hbox{and}\quad f(\tilde{x}-x_0) \le\mu(t_0)+\gamma. $$

(7.62)

It follows from (7.62), (7.52) and (7.55) that

$$ \|x_0-\tilde{x}\| \ge\rho(\tilde{x},A)- \rho(x_0,A)\ge\rho( \tilde{x},A) -t_0 \ge\rho(\tilde{x},A)- r/2, $$

(7.63)

and hence by (6.51),

$$ \|x_0-\tilde{x}\|>r/2. $$

(7.64)

It follows from (7.62) and (7.57) that

$$ \|x_0-\tilde{x}\| \le n/4. $$

(7.65)

There exist $\bar{x} \in\{\alpha x_{0}+(1-\alpha)\tilde{x}: \alpha\in (0,1)\}$ and α ₀∈(0,1) such that

$$ \|\bar{x}-x_0\|=r/2 $$

(7.66)

and

$$ \bar{x}=\alpha_0x_0+(1- \alpha_0)\tilde{x}. $$

(7.67)

By (7.67) and (7.66), $r/2=\|\bar{x}-x_{0}\|=\|\alpha_{0}x_{0}+(1-\alpha_{0})\tilde{x}-x_{0}\|= (1-\alpha_{0})\|\tilde{x}-x_{0}\|$ and

$$ \alpha_0=1-r\bigl(2\|\tilde{x}-x_0 \|\bigr)^{-1}. $$

(7.68)

Relations (7.68) and (7.65) imply that

$$ \alpha_0 \le1-r/(2n/4) =1-2r/n. $$

(7.69)

Set

$$ \tilde{A}=A_{t_0} \cup\{\bar{x}\}. $$

(7.70)

Now we will estimate $f(\tilde{x}-\bar{x})$. By (7.67), (7.65), (7.37), (7.62) and (7.69),

$$\begin{aligned} f(\tilde{x}-\bar{x}) =& f\bigl(\tilde{x} -\bigl(\alpha_0x_0+(1- \alpha_0)\tilde{x}\bigr)\bigr)= f\bigl(\alpha_0(\tilde{x}-x_0)\bigr) \\ \le&\phi_n(\alpha_0)f(\tilde{x}-x_0)\le \phi_n(\alpha_0) \bigl(\mu (t_0)+\gamma \bigr)\\ \le& \phi_n(1-2r/n) \bigl(\mu(t_0)+\gamma\bigr). \end{aligned}$$

Thus

$$ f(\tilde{x}-\bar{x})\le \phi_n(1-2r/n) \bigl( \mu(t_0)+\gamma\bigr) \le\mu(t_0)\phi_n(1-2r/n) +\gamma. $$

(7.71)

By (7.70), (7.53), (7.58) and (7.71), for each $x \in \tilde{A}\setminus\{\bar{x}\} \subset A_{t_{0}}$,

$$ f(\tilde{x}-x) \ge\mu(t_0) >f(\tilde{x}-\bar{x}) $$

(7.72)

and therefore

$$ f(\tilde{x}-\bar{x}) =\rho_f(\tilde{x},\tilde{A}). $$

(7.73)

There exists δ∈(0,δ ₀) such that

$$ 2^{n+4}\delta<\delta_0 $$

(7.74)

and

$$\begin{aligned} &\bigl|f(z)-f(\tilde{x}-\bar{x})\bigr| \le\gamma/4 \\ &\quad\hbox{for all } z \in X \hbox{ satisfying } \bigl\| z-(\tilde{x}-\bar{x})\bigr\| \le2^{n+3}\delta. \end{aligned}$$

(7.75)

By (7.70), (7.40), (7.66), (7.62), (7.55) and (7.52),

$$ \tilde{H}(\tilde{A},A) \le H(\tilde{A},A) \le r. $$

(7.76)

Relations (7.76) and (7.73) imply (7.42). Assume now that

$$ \tilde{y} \in X, \quad\|\tilde{y}-\tilde{x}\| \le\delta $$

(7.77)

and

$$ B \in S(X) \quad\hbox{and}\quad h(\tilde{A},B) \le\delta. $$

(7.78)

First we will show that

$$ \rho_f(\tilde{y},B) \le\mu(t_0) \phi_n(1-2r/n) +2\gamma. $$

(7.79)

By (7.78) and the definition of h (see (7.41)), $h_{n}(\tilde{A},B)(1+h_{n}(\tilde{A},B))^{-1} \le2^{n}\delta$. When combined with (7.74), this inequality implies that

$$ h_n(\tilde{A},B) \le2^n\delta \bigl(1-2^n\delta\bigr)^{-1} \le2^{n+1}\delta. $$

(7.80)

It follows from (7.41) and the definition of n (see (7.57), (7.56)) that $\|\tilde{x}-\bar{x}\| \le n/2$ and $\|\bar{x}\| \le n$. When combined with (7.70) and (7.80), this implies that $\rho(\bar{x}, B) \le2^{n+1}\delta$. Therefore there exists $\bar{y} \in B$ such that $\|\bar{x}-\bar{y}\| \le2^{n+2}\delta$. Combining this inequality with (7.77), we see that $\|(\bar{y}-\tilde{y})-(\bar{x}-\tilde{x})\|\le\|\bar{x}-\bar{y}\|+ \|\tilde{y}-\tilde{x}\|\le2^{n+3}\delta$. It follows from this inequality and (7.75) that $f(\tilde{y} -\bar{y})\le f(\tilde{x}-\bar{x})+ \gamma/4$. By the last inequality and (7.71), $f(\tilde{y} -\bar{y})\le\mu(t_{0})\phi_{n}(1-2r/n) +2\gamma$. This implies (7.79).

Assume now that z∈B satisfies (7.43). To complete the proof of the lemma it is sufficient to show that $\|\bar{x}-z\| \le \varepsilon$. Assume the contrary. Then

$$ \|\bar{x}-z\|>\varepsilon . $$

(7.81)

We will show that there exists $\bar{z} \in\tilde{A}$ such that

$$ \|z-\bar{z}\| \le2^{n+2}\delta. $$

(7.82)

We have already shown that (7.80) holds. By (7.43), (7.79), (7.58) and (7.74),

$$f(\tilde{y} -z) \le\rho_f(\tilde{y},B)+\delta\le \phi_n(1-2r/n) \mu(t_0) +2\gamma+\delta\le\mu(0)+1/2. $$

Hence $\|z-\tilde{y}\| \le n/4$ by (7.57), and by (7.77) and (7.56),

$$\|z\| \le n/4+\|\tilde{y}\| \le n/4+\|\tilde{x}\|+\|\tilde{y} -\tilde{x}\|\le n. $$

Thus ∥z∥≤n. The inclusion z∈B and (7.80) now imply that $\rho(z,\tilde{A}) \le h_{n}(B,\tilde{A}) \le2^{n+1}\delta$. Therefore there exists $\bar{z} \in\tilde{A}$ such that (7.82) holds. It follows from (7.82), (7.81), (7.70), (7.74) and (7.59) that

$$ \bar{z} \in A_{t_0}. $$

(7.83)

By (7.82) and (7.77), $\|z+\tilde{x} -\tilde{y} -\bar{z}\| \le\|\tilde{x} -\tilde{y}\|+\|z-\bar{z}\| \le2^{n+2}\delta+\delta\le2^{n+3}\delta$. It follows from this inequality, (7.83), (7.52) and (7.74) that

$$\rho(z+\tilde{x}-\tilde{y},A) \le\|z+\tilde{x} -\tilde{y} -\bar{z}\|+\rho(\bar{z},A) \le2^{n+3}\delta+t_0 \le t_0+ \delta_0. $$

Thus $z+\tilde{x}-\tilde{y} \in A_{t_{0}+\delta_{0}}$. By this inclusion, (7.52), (7.53) and (7.61),

$$f(\tilde{y}-z)=f\bigl(\tilde{x}-(z+\tilde{x}-\tilde{y})\bigr) \ge \rho_f(\tilde{x}, A_{t_0+\delta_0})=\mu(t_0+ \delta_0) \ge\mu(t_0) -\gamma. $$

Hence, by (7.43), (7.79), (7.59) and (7.74),

$$\begin{aligned} \mu(t_0)-\gamma \le& f(\tilde{y}-z) \le\rho_f(\tilde{y},B)+\delta \le\phi_n(1-2r/n) \mu(t_0) +2\gamma+\delta \\ \le&\phi_n(1-2r/n) \mu(t_0) +3\gamma. \end{aligned}$$

Thus μ(t ₀)−γ≤ϕ _n (1−2r/n)μ(t ₀)+3γ, which contradicts (7.58). This completes the proof of Lemma 7.12. □

7.7 Proofs of Theorems 7.8–7.11

The cornerstone of our proofs is the property established in Lemma 7.12.

By Lemma 7.12, for each (A,x)∈S(X)×X and each integer k≥1, there exist A(x,k)∈S(X), $\bar{x}(A,k) \in A(x,k)$, and δ(x,A,k)>0 such that

$$ \tilde{H}\bigl(A, A(x,k)\bigr) \le2^{-k},\quad\quad f\bigl(x-\bar{x}(A,k)\bigr)=\rho_f\bigl(x,A(x,k)\bigr), $$

(7.84)

and the following property holds:

(P1):

For each y∈X satisfying ∥y−x∥≤2δ(x,A,k), each B∈S(X) satisfying h(B,A(x,k))≤2δ(x,A,k) and each z∈B satisfying f(y−z)≤ρ _f (y,B)+2δ(x,A,k), the inequality $\|z-\bar{x}(A,k)\| \le2^{-k}$ holds.

For each (A,x)∈S(X)×X and each integer k≥1, define

$$\begin{aligned} V(A,x,k) =&\bigl\{ (B,y) \in S(X) \times X: \\ &h\bigl(B,A(x,k)\bigr)<\delta(x,A,k) \hbox{ and } \|y-x\| < \delta(x,A,k)\bigr\} \end{aligned}$$

(7.85)

and

$$ U(A,x,k)=\bigl\{ B \in S(X): h\bigl(B,A(x,k)\bigr)<\delta(x,A,k) \bigr\} . $$

(7.86)

Now set

$$ \varOmega = \bigcap_{n=1}^{\infty} \bigcup\bigl\{ V(A,x,k): (A,x) \in S(X) \times X, k \ge n\bigr\} , $$

(7.87)

and for each x∈X let

$$ \varOmega _x=\bigcap_{n=1}^{\infty} \bigcup\bigl\{ U(A,x,k): A \in S(X), k \ge n\bigr\} . $$

(7.88)

It is easy to see that Ω _x ×{x}⊂Ω for all x∈X, Ω _x is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of S(X) for all x∈X, and Ω is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of S(X)×X.

Completion of the proof of Theorem 7.9

Let $(A,\tilde{x}) \in \varOmega$. We will show that $(A,\tilde{x})$ has property (C2). By the definition of Ω (see (7.87)), for each integer n≥1, there exist an integer k _n ≥n and a pair (A _n ,x _n )∈S(X)×X such that

$$ (A,\tilde{x}) \in V(A_n,x_n,k_n). $$

(7.89)

Let $\{z_{i}\}_{i=1}^{\infty} \subset A$ be such that

$$ \lim_{i \to\infty} f(\tilde{x}-z_i)= \rho_f(\tilde{x}, A). $$

(7.90)

Fix an integer n≥1. It follows from (7.89), (7.85) and property (P1) that for all large enough integers i,

$$f(\tilde{x}-z_i) < \rho_f(\tilde{x},A)+ \delta(x_n,A_n,k_n) $$

and

$$\bigl\| z_i-\bar{x}_n(A_n,k_n)\bigr\| \le2^{-n}. $$

Since n≥1 is arbitrary, we conclude that $\{z_{i}\}_{i=1}^{\infty}$ is a Cauchy sequence which converges to some $\tilde{y} \in A$. Clearly $f(\tilde{x}-\tilde{y})=\rho_{f}(\tilde{x}, A)$. If the minimizer $\tilde{y}$ were not unique we would be able to construct a nonconvergent minimizing sequence $\{z_{i}\}_{i=1}^{\infty}$. Thus $\tilde{y}$ is the unique solution to problem (P) (with $x= \tilde{x}$).

Let ε>0 be given. Choose an integer n>4/min{1,ε}. By property (P1), (7.89) and (7.85),

$$ \bigl\| \tilde{y}-\bar{x}_n(A_n,k_n)\bigr\| \le2^{-n}. $$

(7.91)

Assume that z∈X satisfies $\|z-\tilde{x}\| \le \delta(x_{n},A_{n},k_{n})$, B∈S(X) satisfies h(A,B)≤δ(x _n ,A _n ,k _n ), and y∈B satisfies f(z−y)≤ρ _f (z,B)+δ(x _n ,A _n ,k _n ). Then

$$h\bigl(B,A_n(x_n,k_n)\bigr) \le2 \delta(x_n,A_n,k_n) \quad\hbox{and}\quad \bigl\| z-\bar{x}_n(A_n,k_n)\bigr\| \le2\delta(x_n,A_n,k_n) $$

by (7.89) and (7.85). Now it follows from property (P1) that

$$\bigl\| y-\bar{x}_n(A_n,k_n)\bigr\| \le2^{-n}. $$

When combined with (7.91), this implies that

$$\|y-\tilde{y}\| \le2^{1-n} <\varepsilon . $$

The proof of Theorem 7.9 is complete. □

Theorem 7.8 follows from Theorem 7.9 and the inclusion $\varOmega _{\tilde{x}} \times\{\tilde{x}\} \subset \varOmega$.

Although a variant of Theorem 7.10 also follows from Theorem 7.9 by a classical result of Kuratowski and Ulam [87], the following direct proof may also be of interest.

Proof of Theorem 7.10

Let the sequence $\{x_{i}\}_{i=1}^{\infty} \subset X_{0}$ be everywhere dense in X ₀. Set ${\mathcal{F}}=\bigcap_{p=1}^{\infty} \varOmega _{x_{p}}$. Clearly, $\mathcal{F}$ is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of S(X).

Let $A \in{ \mathcal{F}}$ and let p,n≥1 be integers. Clearly, $A \in \varOmega _{x_{p}}$ and by (7.88) and (7.86), there exist A _n ∈S(X) and an integer k _n ≥n such that

$$ h\bigl(A,A_n(x_p,k_n)\bigr)< \delta(x_p,A_n,k_n) \quad\hbox{with } A \in S(X). $$

(7.92)

It follows from this inequality and property (P1) that the following property holds:

(P2):

For each y∈X satisfying ∥y−x _p ∥≤δ(x _p ,A _n ,k _n ) and each z∈A satisfying f(y−z)≤ρ _f (y,A)+2δ(x _p ,A _n ,k _n ), the inequality $\|z-\bar{x}_{p}(A_{n},k_{n})\| \le2^{-n}$ holds.

Set W(p,n)={z∈X ₀:∥z−x _p ∥<δ(x _p ,A _n ,k _n )} and

$$F=\bigcap_{n=1}^{\infty} \bigcup\bigl\{ W(p,n): p=1,2,\dots\bigr\} . $$

It is clear that F is a countable intersection of open and everywhere dense subsets of X ₀.

Let x∈F be given. Consider a sequence $\{z_{i}\}_{i=1}^{\infty} \subset A$ such that

$$ \lim_{i \to\infty}f(x-z_i)= \rho_f(x,A). $$

(7.93)

Let ε>0. Choose an integer n>8/min{1,ε}. There exists an integer p≥1 such that x∈W(p,n). By the definition of W(p,n), ∥x−x _p ∥<δ(x _p ,A _n ,k _n ). It follows from this inequality, (7.93) and property (P2) that for all sufficiently large integers i, f(x−z _i )≤ρ _f (x,A)+δ(x _p ,A _n ,k _n ) and $\|z_{i}-\bar{x}_{p}(A_{n},k_{n})\| \le2^{-n}<\varepsilon$. Since ε>0 is arbitrary, we conclude that $\{z_{i}\}_{i=1}^{\infty}$ is a Cauchy sequence which converges to $\tilde{y} \in A$. Clearly, $\tilde{y}$ is the unique minimizer of the minimization problem z→f(x−z), z∈A. Note that we have shown that any sequence $\{z_{i}\}_{i=1}^{\infty} \subset A$ satisfying (7.93) converges to $\tilde{y}$. This completes the proof of Theorem 7.10. □

7.8 A Porosity Result in Best Approximation Theory

Let D be a nonempty compact subset of a complete hyperbolic space (X,ρ,M) and denote by S(X) the family of all nonempty closed subsets of X. We endow S(X) with a pair of natural complete metrics and show that there exists a set Ω⊂S(X) such that its complement S(X)∖Ω is σ-porous with respect to this pair of metrics and such that for each A∈Ω and each $\tilde{x} \in D$, the following property holds: the set $\{y \in A: \rho(\tilde{x},y)=\rho(\tilde{x},A)\}$ is nonempty and compact, and each sequence $\{y_{i}\}_{i=1}^{\infty} \subset A$ which satisfies $\lim_{i \to\infty} \rho(\tilde{x},y_{i})=\rho(\tilde{x},A)$ has a convergent subsequence. This result was obtained in [147].

Let (X,ρ,M) be a complete hyperbolic space. For each x∈X and each A⊂X, set

$$\rho(x,A)=\inf\bigl\{ \rho(x,y): y \in A\bigr\} . $$

Denote by S(X) the family of all nonempty closed subsets of X. For each A,B∈S(X), define

$$ H(A,B):=\max\bigl\{ \sup\bigl\{ \rho(x,B): x \in A\bigr\} , \sup\bigl\{ \rho(y,A): y \in B\bigr\} \bigr\} $$

(7.94)

and

$$\tilde{H}(A,B):=H(A,B) \bigl(1+H(A,B)\bigr)^{-1}. $$

Here we use the convention that ∞/∞=1. It is easy to see that $\tilde{H}$ is a metric on S(X) and that the metric space $(S(X),\tilde{H})$ is complete.

Fix θ∈X. For each natural number n and each A,B∈S(X), we set

$$ h_n(A,B)=\sup\bigl\{ \bigl|\rho(x,A)-\rho(x,B)\bigr|: x \in X \hbox{ and } \rho(x,\theta) \le n\bigr\} $$

(7.95)

and

$$h(A,B)=\sum_{n=1}^{\infty}\bigl[2^{-n}h_n(A,B) \bigl(1+h_n(A,B)\bigr)^{-1}\bigr]. $$

Once again, it is not difficult to see that h is a metric on S(X) and that the metric space (S(X),h) is complete. Clearly,

$$\tilde{H}(A,B) \ge h(A,B) \quad\hbox{for all } A,B \in S(X). $$

We equip the set S(X) with the pair of metrics $\tilde{H}$ and h and prove the following theorem which is the main result of [147].

Theorem 7.13

Given a nonempty compact subset D of a complete hyperbolic space (X,ρ,M), there exists a set Ω⊂S(X) such that its complement S(X)∖Ω is σ-porous with respect to the pair of metrics $(h,\tilde{H})$, and such that for each A∈Ω and each $\tilde{x} \in D$, the following property holds:

The set $\{y \in A: \rho(\tilde{x},y)=\rho(\tilde{x},A)\}$ is nonempty and compact and each sequence $\{y_{i}\}_{i=1}^{\infty} \subset A$ which satisfies $\lim_{i \to\infty} \rho(\tilde{x},y_{i})=\rho(\tilde{x},A)$ has a convergent subsequence.

7.9 Two Lemmata

Let (X,ρ,M) be a complete hyperbolic space and let D be a nonempty compact subset of X. In the proof of Theorem 7.13 we will use the following two lemmata.

Lemma 7.14

Let q be a natural number, A∈S(X), ε∈(0,1), r∈(0,1], and let Q={ξ ₁,…,ξ _q } be a finite subset of D. Then there exists a finite set $\{\tilde{\xi}_{1},\dots,\tilde{\xi}_{q}\} \subset X$ such that

$$ \rho(\tilde{\xi}_i,A) \le r,\quad i=1,\ldots,q, $$

(7.96)

and such that the set $\tilde{A}:=A\cup\{\tilde{\xi}_{1},\dots,\tilde{\xi}_{q}\}$ has the following properties:

$$ \rho\bigl(\xi_i,\{\tilde{\xi}_1,\ldots, \tilde{\xi}_q\}\bigr)=\rho(\xi_i, \tilde{A}),\quad i=1,\ldots,q; $$

(7.97)

(P3):

if i∈{1,…,q}, $x \in\tilde{A}$, and $\rho(\xi_{i},x) \le \rho(\xi_{i},\tilde{A})+\varepsilon r/4$, then

$$\rho\bigl(x,\{\tilde{\xi}_1,\dots,\tilde{\xi}_q\}\bigr) \le \varepsilon . $$

Proof

Let i∈{1,…,q}. There are two cases: (1) ρ(ξ _i ,A)≤r; (2) ρ(ξ _i ,A)>r. In the first case we set

$$ \tilde{\xi}_i=\xi_i. $$

(7.98)

In the second case, we first choose x _i ∈A for which

$$ \rho(\xi_i,x_i) \le\rho( \xi_i,A)+r/4, $$

(7.99)

and then choose

$$ \tilde{\xi}_i \in\bigl\{ \gamma x_i \oplus(1-\gamma)\xi_i: \gamma\in (0,1)\bigr\} $$

(7.100)

such that

$$ \rho(\tilde{\xi}_i,x_i)=r \quad\hbox{and}\quad \rho(\tilde{\xi}_i,\xi_i)=\rho(x_i, \xi_i)-r. $$

(7.101)

Clearly, (7.96) holds. Consider now the set $\tilde{A}=A \cup\{\tilde{\xi}_{1},\dots,\tilde{\xi}_{q}\}$.

Let i∈{1,…,q}. It is not difficult to see that if ρ(ξ _i ,A)≤r, then the assertion of the lemma is true. Consider the case where ρ(ξ _i ,A)>r. It follows from (7.99) and (7.101) that

$$\begin{aligned} \rho\bigl(\xi_i,\{\tilde{\xi}_1,\dots, \tilde{\xi}_q\}\bigr) \le& \rho(\xi_i,\tilde{\xi}_i) = \rho(x_i,\xi_i)-r \\ \le&\rho(\xi_i,A)+r/4-r=\rho(\xi_i,A)-3r/4. \end{aligned}$$

Therefore

$$\rho\bigl(\xi_i,\{\tilde{\xi}_1,\dots,\tilde{\xi}_q\}\bigr)=\rho(\xi_i, \tilde{A}), $$

and if $x \in\tilde{A}$ and $\rho(\xi_{i},x) \le\rho(\xi_{i},\tilde{A})+r/2$, then $x \in\{\tilde{\xi}_{1}, \dots, \tilde{\xi}_{q}\}$. This completes the proof of Lemma 7.14. □

For each ε∈(0,1) and each natural number n, choose a number

$$ \alpha(\varepsilon ,n) \in\bigl(0,16^{-n-2}\varepsilon \bigr) $$

(7.102)

and a natural number n ₀ such that

$$ \rho(x,\theta) \le n_0, \quad x \in D. $$

(7.103)

Lemma 7.15

Let n≥n ₀ be a natural number, A∈S(X), ε∈(0,1), r∈(0,1], and

$$ \alpha=\alpha(\varepsilon ,n). $$

(7.104)

Assume that

$$ \bigl\{ z \in A: \rho(z,\theta) \le n\bigr\} \neq \emptyset. $$

(7.105)

Then there exist a natural number q and a finite set $\{\tilde{\xi}_{1},\dots,\tilde{\xi}_{q}\} \subset X$ such that

$$ \rho(\tilde{\xi}_i,A) \le r,\quad i=1,\dots,q, $$

(7.106)

and if $\tilde{A}:=A \cup\{\tilde{\xi}_{1},\dots,\tilde{\xi}_{q}\}$, u∈D, B∈S(X),

$$ h(\tilde{A},B) \le\alpha r, $$

(7.107)

and

$$ z \in B, \quad\rho(u,z) \le\rho(u,B)+\varepsilon r/16, $$

(7.108)

then

$$ \rho\bigl(z,\{\tilde{\xi}_1,\dots,\tilde{\xi}_q\}\bigr) \le \varepsilon . $$

(7.109)

Proof

Since D is compact, there are a natural number q and a finite subset {ξ ₁,…,ξ _q } of D such that

$$ D \subset \bigcup_{i=1}^q \bigl\{ z \in X: \rho(z,\xi_i) <\alpha r\bigr\} . $$

(7.110)

By Lemma 7.14, there exists a finite set $\{\tilde{\xi}_{1},\dots\tilde{\xi}_{q}\} \subset X$ such that (7.106) holds, and the set $\tilde{A}:=A\cup\{\tilde{\xi}_{1},\dots,\tilde{\xi}_{q}\}$ satisfies (7.97) and has the following property:

(P4):

If i∈{1,…,q}, $x \in\tilde{A}$, and $\rho(\xi_{i},x) \le \rho(\xi_{i},\tilde{A})+\varepsilon r/8$, then

$$\rho\bigl(x,\{\tilde{\xi}_1,\dots,\tilde{\xi}_q\}\bigr) \le \varepsilon /2. $$

Assume that u∈D, B∈S(X), and that (7.107) holds. By (7.110), there is j∈{1,…,q} such that

$$ \rho(\xi_j,u) <\alpha r. $$

(7.111)

We will show that

$$ \rho(u,B) <\rho(\xi_j,\tilde{A})+4 \cdot16^n \alpha r. $$

(7.112)

Indeed, there exists p∈{1,…,q} such that

$$\rho(\xi_j,\tilde{\xi}_p)=\rho\bigl(\xi_j, \{\tilde{\xi}_1,\dots,\tilde{\xi}_q\}\bigr). $$

By (7.97),

$$ \rho(\xi_j,\tilde{\xi}_p)=\rho( \xi_j,\tilde{A}). $$

(7.113)

By (7.111),

$$ \bigl|\rho(u,\tilde{A})-\rho(\xi_j,\tilde{A})\bigr| \le\alpha r. $$

(7.114)

When combined with (7.113), this inequality implies that

$$ \bigl|\rho(u,\tilde{A})-\rho(\xi_j,\tilde{\xi}_p)\bigr| \le\alpha r. $$

(7.115)

Now (7.113), (7.105) and (7.103) imply that

$$ \rho(\xi_j,\tilde{\xi}_p) \le\rho( \xi_j,A) \le2n \quad\hbox{and}\quad \rho(\tilde{\xi}_p,\theta) \le3n. $$

(7.116)

It follows from (7.95) and (7.107) that

$$h_{4n}(\tilde{A},B) \bigl(1+h_{4n}(\tilde{A},B) \bigr)^{-1} \le2^{4n}h(\tilde{A},B) \le 2^{4n} \alpha r, $$

and when combined with (7.104) and (7.102), this inequality yields

$$ h_{4n}(\tilde{A},B) \le2^{4n}\alpha r \bigl(1-2^{4n}\alpha r\bigr)^{-1} <2^{4n+1}\alpha r. $$

(7.117)

Since $\tilde{\xi}_{p} \in\tilde{A}$, it follows from (7.117), (7.116) and (7.97) that $\rho({\tilde{\xi}}_{p},B) <2^{4n+1}\alpha r$ and there exists v∈X such that

$$ v \in B \quad\hbox{and}\quad \rho(\tilde{\xi}_p,v) <2\alpha r 16^n. $$

(7.118)

By (7.118), (7.111), (7.113) and (7.118),

$$\begin{aligned} \rho(u,B) \le&\rho(u,v) \le\rho(u,\tilde{\xi}_p)+\rho(\tilde{\xi}_p,v) \le \rho(u,\xi_j)+\rho(\xi_j,\tilde{\xi}_p)+ \rho(\tilde{\xi}_p,v)\\ <&\alpha r+ \rho(\xi_j,\tilde{A})+2 \cdot16^n\alpha r. \end{aligned}$$

Hence (7.112) is valid.

Now let (7.108) hold. Then by (7.108), (7.112) and (7.102),

$$\begin{aligned} \rho(z,u) \le&\rho(u,B)+\varepsilon r/16 <\rho(\xi_j, \tilde{A})+4\cdot 16^n \alpha r+\varepsilon r/16 \\ <&\rho(\xi_j,\tilde{A})+\varepsilon r/8. \end{aligned}$$

(7.119)

Therefore (7.119) and (7.116) imply that

$$\rho(z,u) \le\rho(\xi_j,\tilde{A})+\varepsilon r/8 \le2n+r/8. $$

It follows from this inequality, (7.111) and (7.103) that

$$\begin{aligned} \rho(z,\theta) \le&\rho(z,u)+\rho(u,\theta) \le2n+r/8+\rho (u,\theta) \\ \le& 2n+r/8+\rho(u,\xi_j)+\rho(\xi_j,\theta) \le2n+r/8+\alpha r+n \le4n. \end{aligned}$$

Since z∈B, it follows from (7.97) and (7.117) that

$$\rho(z,\tilde{A})=\bigl|\rho(z,\tilde{A})-\rho(z,B)\bigr| \le h_{4n}(\tilde{A},B)< 2 \cdot16^n\alpha r. $$

Therefore there exists $\tilde{z} \in\tilde{A}$ such that

$$ \rho(z,\tilde{z})<2\cdot16^n \alpha r. $$

(7.120)

By (7.111), (7.120), (7.108), (7.112) and (7.102),

$$\begin{aligned} \rho(\tilde{z},\xi_j) \le&\rho(\xi_j,u)+\rho(u,z)+ \rho(z,\tilde{z})< \alpha r+ \rho(u,z)+2 \cdot16^n \alpha r \\ \le&\alpha r+2 \cdot16^n \alpha r+\rho(u,B)+\varepsilon r/16\\ <&\varepsilon r /16+\alpha r+2 \cdot16^n \alpha r+ \rho(\xi_j,\tilde{A})+4 \cdot16^n \alpha r \\ \le&\rho(\xi_j, \tilde{A})+ 8\cdot16^n \alpha r+ \varepsilon r/16 \le\rho(\xi_j,\tilde{A})+\varepsilon r/8 \end{aligned}$$

and

$$ \rho(\tilde{z}, \xi_j) <\rho(\xi_j, \tilde{A})+\varepsilon r/8. $$

(7.121)

Since $\tilde{z} \in\tilde{A}$, it follows from (7.121) and property (P4) that $\rho(\tilde{z},\{\tilde{\xi}_{1},\dots,\tilde{\xi}_{q}\}) \le \varepsilon /2$. When combined with (7.120) and (7.102), this inequality implies that

$$\rho\bigl(z,\{\tilde{\xi}_1,\dots,\tilde{\xi}_q\}\bigr) \le \varepsilon . $$

This completes the proof of Lemma 7.15. □

7.10 Proof of Theorem 7.13

For each integer k≥1, denote by Ω _k the set of all A∈S(X) which have the following property:

(P5):

There exist a nonempty finite set Q⊂X and a number δ>0 such that if u∈D, x∈A and ρ(u,x)≤ρ(u,A)+δ, then ρ(x,Q)≤1/k.

It is clear that Ω _k+1⊂Ω _k , k=1,2,…. Set $\varOmega =\bigcap_{k=1}^{\infty} \varOmega _{k}$.

Let k≥n ₀ (see (7.103)) be an integer. We will show that S(X)∖Ω _k is σ-porous with respect to the pair $(h,\tilde{H})$. For any integer n≥k, define

$$E_{nk}=\bigl\{ A \in S(X) \setminus \varOmega _k: \bigl\{ z \in A: \rho(z,\theta) \le n\bigr\} \neq\emptyset\bigr\} . $$

By Lemma 7.15, E _nk is porous with respect to the pair $(h,\tilde{H})$ for all integers n≥k. Thus $S(X)\setminus \varOmega _{k}= \bigcup_{n=k}^{\infty}E_{nk}$ is σ-porous with respect to $(h,\tilde{H})$. Hence $S(X) \setminus \varOmega =\bigcup_{k=n_{0}}^{\infty}(S(X)\setminus \varOmega _{k})$ is also σ-porous with respect to the pair of metrics $(h,\tilde{H})$.

Let A∈Ω. Since A∈Ω _k for each integer k≥1, it follows from property (P5) that for any integer k≥1, there exist a nonempty finite set Q _k ⊂X and a number δ _k >0 such that the following property also holds:

(P6):

If u∈D, x∈A, and ρ(u,x)≤ρ(x,A)+δ _k , then ρ(x,Q _k )≤1/k.

Let u∈D. Consider a sequence $\{x_{i}\}_{i=1}^{\infty} \subset A$ such that lim _i→∞ ρ(u,x _i )=ρ(u,D). By property (P6), for each integer k≥1, there exists a subsequence $\{x_{i}^{(k)}\}_{i=1}^{\infty}$ of $\{x_{i}\}_{i=1}^{\infty}$ such that the following two properties hold:

• (i) $\{x_{i}^{(k+1)}\}_{i=1}^{\infty}$ is a subsequence of $\{x_{i}^{(k)}\}_{i=1}^{\infty}$ for all integers k≥1;

• (ii) for any integer k≥1, $\rho(x_{j}^{(k)},x_{s}^{(k)}) \le2/k$ for all integers j,s≥1.

These properties imply that there exists a subsequence $\{x_{i}^{*}\}_{i=1}^{\infty}$ of $\{x_{i}\}_{i=1}^{\infty}$ which is a Cauchy sequence. Therefore $\{x_{i}^{*}\}_{i=1}^{\infty}$ converges to a point $\tilde{x} \in A$ which satisfies $\rho(\tilde{x},u)=\lim_{i \to \infty}\rho(x_{i},u)=\rho(u,D)$. This completes the proof of Theorem 7.13.

7.11 Porous Sets and Generalized Best Approximation Problems

Given a closed subset A of a Banach space X, a point x∈X and a Lipschitzian (on bounded sets) function f:X→R ¹, we consider the problem of finding a solution to the minimization problem min{f(x−y):y∈A}. For a fixed function f, we define an appropriate complete metric space $\mathcal{M}$ of all pairs (A,x) and construct a subset Ω of $\mathcal{M}$, with a σ-porous complement ${\mathcal{M}} \setminus \varOmega$, such that for each pair in Ω, our minimization problem is well posed.

Let (X,∥⋅∥) be a Banach space and let f:X→R ¹ be a Lipschitzian (on bounded sets) function. Assume that

$$\begin{gathered} \inf\bigl\{ f(x): x \in X\bigr\} \hbox{ is attained at a unique point } x_* \in X, \end{gathered}$$

(7.122)

$$\begin{gathered} \lim_{\|u\| \to\infty}f(u)=\infty, \end{gathered}$$

(7.123)

$$\begin{gathered} \hbox{if } \{x_i\}_{i=1}^{\infty} \subset X \hbox{ and } \lim_{i \to\infty}f(x_i)=f(x_*), \hbox{ then }\lim_{i \to\infty} x_i=x_*, \end{gathered}$$

(7.124)

$$\begin{gathered} \begin{aligned}[b] & f\bigl(\alpha x+(1-\alpha)x_*\bigr) \le\alpha f(x) +(1- \alpha)f(x_*) \\ &\quad\hbox{for all } x \in X \hbox { and all } \alpha\in(0,1), \end{aligned} \end{gathered}$$

(7.125)

and that for each natural number n, there exists k _n >0 such that

$$ \bigl|f(x)-f(y)\bigr| \le k_n \|x-y\| \quad\hbox{for each } x,y \in X \hbox{ satisfying } \|x\|,\|y\| \le n. $$

(7.126)

Clearly, (7.125) holds if f is convex.

Given a closed subset A of X and a point x∈X, we consider the minimization problem

$$ \min\bigl\{ f(x-y): y \in A\bigr\} . $$

(P)

For each x∈X and each A⊂X, set

$$\rho(x,A)=\inf\bigl\{ \|x-y\|: y \in A\bigr\} $$

and

$$\rho_f(x,A)=\inf\bigl\{ f(x-y): y \in A\bigr\} . $$

Denote by S(X) the collection of all nonempty closed subsets of X. For each A,B∈S(X), define

$$ H(A,B):=\max\bigl\{ \sup\bigl\{ \rho(x,B): x \in A\bigr\} , \sup \bigl\{ \rho(y,A): y \in B\bigr\} \bigr\} $$

(7.127)

and

$$\tilde{H}(A,B):=H(A,B) \bigl(1+H(A,B)\bigr)^{-1}. $$

Here we use the convention that ∞/∞=1.

It is not difficult to see that the metric space $(S(X),\tilde{H})$ is complete.

For each natural number n and each A,B∈S(X), we set

$$ h_n(A,B):=\sup\bigl\{ \bigl|\rho(x,A)-\rho(x,B)\bigr|: x \in X \hbox{ and } \|x\| \le n\bigr\} $$

(7.128)

and

$$h(A,B):=\sum_{n=1}^{\infty}\bigl[2^{-n}h_n(A,B) \bigl(1+h_n(A,B)\bigr)^{-1}\bigr]. $$

Once again, it is not difficult to see that h is a metric on S(X) and that the metric space (S(X),h) is complete. Clearly, $\tilde{H}(A,B) \ge h(A,B)$ for all A,B∈S(X).

We equip the set S(X) with the pair of metrics $\tilde{H}$ and h. The topologies induced by the metrics $\tilde{H}$ and h on S(X) will be called the strong topology and the weak topology, respectively.

Let A∈S(X) and $\tilde{x} \in X$ be given. We say that the best approximation problem

$$f(\tilde{x}-y) \to\min,\quad y \in A, $$

is strongly well posed if there exists a unique $\bar{x} \in A$ such that

$$f(\tilde{x}-\bar{x})=\inf\bigl\{ f(\tilde{x}-y): y \in A\bigr\} $$

and the following property holds:

For each ε>0, there exists δ>0 such that if z∈X satisfies $\|z-\tilde{x}\| \le\delta$, B∈S(X) satisfies h(A,B)≤δ, and y∈B satisfies f(z−y)≤ρ _f (z,B)+δ, then $\|y-\bar{x} \| \le \varepsilon$.

We now state four results obtained in [151]. Their proofs will be given in the next sections.

Theorem 7.16

Let $\tilde{x} \in X$ be given. Then there exists a set Ω⊂S(X) such that its complement S(X)∖Ω is σ-porous with respect to $(h,\tilde{H})$ and for each A∈Ω, the problem $f(\tilde{x}-y) \to\min$, y∈A, is strongly well posed.

To state our second result, we endow the Cartesian product S(X)×X with the pair of metrics d ₁ and d ₂ defined by

$$\begin{aligned} &d_1\bigl((A,x),(B,y)\bigr)=h(A,B)+\|x-y\|, \\ &d_2 \bigl((A,x),(B,y)\bigr)=\tilde{H}(A,B)+\|x-y\|, \quad x,y \in X, A,B \in S(X). \end{aligned}$$

We will refer to the metrics induced on S(X)×X by d ₂ and d ₁ as the strong and weak metrics, respectively.

Theorem 7.17

There exists a set Ω⊂S(X)×X such that its complement (S(X)×X)∖Ω is σ-porous with respect to (d ₁,d ₂) and for each $(A, \tilde{x}) \in \varOmega$, the minimization problem

$$f(\tilde{x}-y) \to\min,\quad y \in A, $$

is strongly well posed.

In most classical generic results the set A was fixed and x varied in a dense G _δ subset of X. In our first two results the set A is also variable. However, our third result shows that for every fixed A in a subset of S(X) which has a σ-porous complement, the set of all x∈X for which problem (P) is strongly well posed contains a dense G _δ subset of X.

Theorem 7.18

Assume that X ₀ is a closed separable subset of X. Then there exists a set ${\mathcal{F}} \subset S(X)$ such that its complement $S(X) \setminus{ \mathcal{F}}$ is σ-porous with respect to $(h,\tilde{H})$ and for each $A \in {\mathcal{F}}$, the following property holds:

There exists a set F⊂X ₀, which is a countable intersection of open and everywhere dense subsets of X ₀ with the relative topology, such that for each $\tilde{x} \in F$, the minimization problem

$$f(\tilde{x}-y) \to\min,\quad y \in A, $$

is strongly well posed.

Now we will show that Theorem 7.16 implies the following result.

Theorem 7.19

Assume that g:X→R ¹ is a convex function which is Lipschitzian on bounded subsets of X and that inf{g(x):x∈X} is attained at a unique point y _∗∈X, lim_{∥u∥→∞} g(u)=∞, and if $\{y_{i}\}_{i=1}^{\infty} \subset X$ and lim _i→∞ g(y _i )=g(y _∗), then y _i →y _∗ as i→∞. Then there exists a set Ω⊂S(X) such that its complement S(X)∖Ω is σ-porous with respect to $(h,\tilde{H})$ and for each A∈Ω, the following property holds:

There is a unique y _A ∈A such that g(y _A )=inf{g(y):y∈A}. Moreover, for each ε>0, there exists δ>0 such that if y∈A satisfies g(y)≤g(y _A )+δ, then ∥y−y _A ∥≤ε.

Proof

Define f(x)=g(−x), x∈X. It is clear that f is convex and satisfies (7.122)–(7.126). Therefore Theorem 7.16 is valid with $\tilde{x}=0$ and there exists a set Ω⊂S(X) such that its complement S(X)∖Ω is σ-porous with respect to $(h, \tilde{H})$ and for each A∈Ω, the following property holds:

There is a unique $\tilde{y} \in A$ such that

$$g(\tilde{y})=f(-\tilde{y})=\inf\bigl\{ f(-y): y \in A\bigr\} = \inf\bigl\{ g(y): y \in A\bigr\} . $$

Moreover, for each ε>0, there exists δ>0 such that if B∈S(X) satisfies h(A,B)≤δ and x∈B satisfies

$$g(x)=f(-x) \le\rho_f(0,B)+\delta=\inf\bigl\{ f(-y): y \in B\bigr\} + \delta=\inf\bigl\{ g(y): y \in B\bigr\} +\delta, $$

then $\|x-\tilde{y}\| \le \varepsilon$. Theorem 7.19 is proved. □

It is easy to see that in the proofs of Theorems 7.16–7.18 we may assume without any loss of generality that inf{f(x):x∈X}=0. It is also not difficult to see that we may assume without loss of generality that x _∗=0. Indeed, instead of the function f(⋅) we can consider f(⋅+x _∗). This new function also satisfies (7.122)–(7.126). Once Theorems 7.16–7.18 are proved for this new function, they will also hold for the original function f because the mapping (A,x)→(A,x+x _∗), (A,x)∈S(X)×X, is an isometry with respect to both metrics d ₁ and d ₂.

7.12 A Basic Lemma

Let m and n be two natural numbers. Choose a number

$$ c_m >\sup \bigl\{ f(u): u \in X \hbox{ and } \|u\| \le2m+4\bigr\} +2 $$

(7.129)

(see (7.126)). By (7.123), there exists a natural number

$$a_m>m+2 $$

such that

$$ \hbox{if } u \in X \hbox{ and } f(u) \le c_m, \hbox{ then } \|u\| \le a_m. $$

(7.130)

By (7.126), there is k _m >1 such that

$$\begin{aligned} &\bigl|f(x)-f(y)\bigr| \le k_m\|x-y\| \\ & \quad\hbox{for each } x,y \in X \hbox{ satisfying } \|x\|,\|y\| \le4a_m+4. \end{aligned}$$

(7.131)

By (7.131), there exists a positive number

$$ \alpha(m,n)<2^{-4a_m-4}16^{-1}n^{-1} $$

(7.132)

such that

$$ \hbox{if } u \in X \hbox{ satisfies } f(u) \le320a_m \alpha(m,n), \hbox{ then } \|u\| \le(4n)^{-1}. $$

(7.133)

Finally, we choose a positive number

$$ \bar{\alpha}(m,n)<\alpha(m,n)\bigl[(k_m+1)^{-1}2^{-4a_m-16} \bigr]. $$

(7.134)

Lemma 7.20

Let

$$ \alpha=\alpha(m,n),\quad\quad \bar{\alpha}=\bar{\alpha}(m,n), $$

(7.135)

A∈S(X), $\tilde{x} \in X$, r∈(0,1], and assume that

$$ \|\tilde{x}\|\le m \quad\textit{and}\quad \bigl\{ z \in X: \|z\|\le m\bigr\} \cap A \neq \emptyset. $$

(7.136)

Then there exists $\bar{x} \in X$ such that

$$ \rho(\bar{x},A) \le r/8 $$

(7.137)

and for the set $\tilde{A}:=A\cup\{\bar{x}\}$, the following property holds:

If

$$\begin{aligned} B \in S(X),\quad h(\tilde{A},B) \le\bar{\alpha}r,& \end{aligned}$$

(7.138)

$$\begin{aligned} \tilde{y} \in X,\quad \|\tilde{y}-\tilde{x}\|\le\bar{\alpha}r,& \end{aligned}$$

(7.139)

and

$$ z \in B,\quad f(\tilde{y}-z) \le\rho_f(\tilde{y},B)+\alpha r, $$

(7.140)

then

$$ h(A,B) \le r $$

(7.141)

and

$$ \|z-\bar{x}\|\le n^{-1}. $$

(7.142)

Proof

First we choose $\bar{x} \in X$. There are two cases: (1) $\rho(\tilde{x},A) \le r/8$; (2) $\rho(\tilde{x},A)>r/8$. If

$$ \rho(\tilde{x},A) \le r/8, $$

(7.143)

then we set

$$ \bar{x}=\tilde{x} \quad\hbox{and}\quad \tilde{A}=A \cup\{\tilde{x}\}. $$

(7.144)

Now consider the second case where

$$ \rho(\tilde{x},A)>r/8. $$

(7.145)

First, choose x ₀∈A such that

$$ f(\tilde{x}-x_0) \le\rho_f(\tilde{x},A)+ \alpha(m,n)r $$

(7.146)

and then choose

$$ \bar{x} \in\bigl\{ \gamma\tilde{x}+(1-\gamma)x_0: \gamma\in(0,1)\bigr\} $$

(7.147)

such that

$$ \|\bar{x}-x_0\|=r/8 \quad\hbox{and}\quad \|\tilde{x}-\bar{x}\|= \|\tilde{x}-x_0\|-r/8. $$

(7.148)

Finally, set

$$ \tilde{A}=A\cup\{\bar{x}\}. $$

(7.149)

Clearly, there is γ∈(0,1) such that

$$ \bar{x}=\gamma\tilde{x}+(1-\gamma)x_0. $$

(7.150)

It is easy to see that in both cases (7.137) holds and

$$ \tilde{H}(A,\tilde{A}) \le H(A,\tilde{A}) \le r/8. $$

(7.151)

Now assume that z∈X satisfies

$$ z \in\tilde{A}\quad \hbox{and}\quad f(\tilde{x}-z) \le\rho_f( \tilde{x},\tilde{A})+8\alpha (m,n)r. $$

(7.152)

We will show that $\|\bar{x}-z\| \le(2n)^{-1}$. First consider case (1). Then by (7.152), (7.144) and (7.149),

$$f(\bar{x}-z)=f(\tilde{x}-z) \le8\alpha(m,n)r. $$

When combined with (7.133), this inequality implies that

$$\|\bar{x}-z\|\le(4n)^{-1}. $$

Now consider case (2). We first estimate $f(\tilde{x}-\bar{x})$. By (7.150) and (7.125) (with x _∗=0 and f(x _∗)=0),

$$\begin{aligned} f(\tilde{x}-\bar{x}) =&f\bigl(\tilde{x}-\gamma\tilde{x}-(1-\gamma)x_0 \bigr) \\ =&f\bigl((1-\gamma) (\tilde{x}-x_0)\bigr) \le(1- \gamma)f(\tilde{x}-x_0). \end{aligned}$$

(7.153)

By (7.136), there is z ₀∈X such that

$$ z_0 \in A \quad\hbox{and} \quad\|z_0\|\le m. $$

(7.154)

Thus (7.146), (7.132), (7.154) and (7.136) imply that

$$\begin{aligned} f(\tilde{x}-x_0) \le&\rho_f(\tilde{x},\tilde{A})+1 \le f( \tilde{x}-z_0)+1 \\ \le&\sup\bigl\{ f(u): u \in X, \|u\| \le2m+1\bigr\} +1<c_m. \end{aligned}$$

(7.155)

Relations (7.155) and (7.130) imply that

$$ \|x_0-\tilde{x}\|\le a_m. $$

(7.156)

It follows from (7.148), (7.150) and (7.156) that

$$\begin{gathered} \begin{aligned} \|\tilde{x}-x_0\|-r/8&=\|\tilde{x}-\bar{x}\|=\bigl\| \tilde{x}-\gamma\tilde{x}-(1-\gamma)x_0\bigr\| \\ &=(1-\gamma)\|\tilde{x}-x_0\|, \end{aligned} \\ \begin{aligned} 1-\gamma&=\bigl(\|\tilde{x}-x_0\|-r/8\bigr)\|\tilde{x}-x_0 \|^{-1}=1- r\bigl(8\|\tilde{x}-x_0\|\bigr)^{-1} \\ &\le1-r(8a_m)^{-1} \end{aligned} \end{gathered}$$

and that

$$ 1-\gamma\le1-r(8a_m)^{-1}. $$

(7.157)

By (7.153) and (7.157),

$$ f(\tilde{x}-\bar{x}) =(1-\gamma)f(\tilde{x}-x_0) \le \bigl(1-r(8a_m)^{-1}\bigr) f(\tilde{x}-x_0). $$

(7.158)

Relations (7.152) and (7.158) now imply that

$$ f(\tilde{x}-z) \le f(\tilde{x}-\bar{x})+8\alpha r \le8\alpha r+ \bigl(1-r(8a_m)^{-1}\bigr) f(\tilde{x}-x_0). $$

(7.159)

There are two cases:

$$ f(\tilde{x}-x_0) \ge8\cdot18 \alpha a_m $$

(7.160)

and

$$ f(\tilde{x}-x_0) \le8 \cdot18 \alpha a_m . $$

(7.161)

Assume that (7.160) holds. Then it follows from (7.159), (7.146) and (7.160) that

$$\begin{aligned} f(\tilde{x}-z) \le&8\alpha r +f(\tilde{x}-x_0)-r(8a_m)^{-1}f( \tilde{x}-x_0) \\ \le&8\alpha r+\rho_f(\tilde{x},A)+\alpha r-8^{-1} \cdot18 \alpha r <\rho_f (\tilde{x},A). \end{aligned}$$

Thus z∉A and by (7.152) and (7.149),

$$ z=\bar{x}. $$

(7.162)

Now assume that (7.161) is true. By (7.161) and (7.152),

$$f(\tilde{x}-z) \le f(\tilde{x}-x_0) +8\alpha r \le8 \cdot18 \alpha a_m+8\alpha \le160 \alpha a_m. $$

When combined with (7.133), (7.148) and (7.161), this estimate implies that

$$\begin{gathered} \|\tilde{x}-z\| \le(4n)^{-1},\quad\quad \|\tilde{x}-x_0\| \le(4n)^{-1}, \\ \|\tilde{x}-\bar{x}\| <\|\tilde{x}-x_0\|<(4n)^{-1}, \end{gathered}$$

and

$$\|\bar{x}-z\|<(2n)^{-1}. $$

Thus in both cases,

$$\|\bar{x}-z\|<(2n)^{-1}. $$

In other words, we have shown that the following property holds:

(P1):

If z∈X satisfies (7.152), then $\|\bar{x}-z\| \le(2n)^{-1}$.

Now assume that (7.138)–(7.140) hold. By (7.136) and (7.139), we have

$$ \|\tilde{x}\| \le m \quad\hbox{and}\quad \|\tilde{y}\| \le m+1. $$

(7.163)

Relation (7.136) implies that there is z ₀∈X such that

$$ z_0 \in A \quad\hbox{and}\quad \|z_0\| \le m. $$

(7.164)

It follows from (7.128), (7.138), (7.164), (7.134) and (7.128) that

$$ \begin{gathered} h_{4a_m+4}(\tilde{A},B) \bigl(1+h_{4a_m+4}(\tilde{A},B) \bigr)^{-1} \le2^{4a_m+4}h(\tilde{A},B) \le2^{4a_m+4}\bar{\alpha}r, \\ h_{4a_m+4}(\tilde{A},B) \le2^{4a_m+4}\bar{\alpha}r \bigl(1-2^{4a_m+4}\bar{\alpha}r\bigr) \le2^{4a_m+5} \bar{\alpha}r \end{gathered} $$

(7.165)

and

$$\begin{aligned} \rho(z_0,B) \le&\rho(z_0,\tilde{A})+\bigl| \rho(z_0,B)-\rho(z_0,\tilde{A})\bigr| \\ \le& h_{4a_m+4}( \tilde{A},B) \le2^{4a_m+5}\bar{\alpha}r. \end{aligned}$$

(7.166)

Inequalities (7.166), (7.134) and (7.132) imply that ρ(z ₀,B)<1, and that there is $\tilde{z}_{0} \in X$ such that

$$ \tilde{z}_0 \in B \quad\hbox{and}\quad \|\tilde{z}_0-z_0\|<1. $$

(7.167)

Clearly, by (7.164) and (7.167),

$$ \|\tilde{z}_0\|<m+1. $$

(7.168)

Let

$$ \bigl\{ (L,l)\bigr\} \in\bigl\{ (\tilde{A},\tilde{x}),(B,\tilde{y}) \bigr\} . $$

(7.169)

By (7.136), (7.163), (7.164), (7.168) and (7.167),

$$ \|l\| \le m+1 $$

(7.170)

and there is $\bar{u} \in X$ such that

$$ \bar{u} \in L \quad\hbox{and}\quad \|\bar{u}\| \le m+1. $$

(7.171)

Relations (7.171), (7.170) and (7.129) imply that

$$ \rho_f(l,L) \le f(l-\bar{u}) \le\sup\bigl\{ f(u): u \in X, \|u\| \le 2m+2\bigr\} \le c_m-2. $$

(7.172)

Also, relations (7.172), (7.130) and (7.170) imply the following property:

(P2):

If u∈L and f(l−u)≤ρ _f (l,L)+2, then ∥l−u∥≤a _m and ∥u∥≤∥l∥+a _m ≤2a _m .

Now assume that L _i ∈S(X) and l _i ∈X, i=1,2, satisfy

$$ \bigl\{ (L_1,l_1),(L_2,l_2) \bigr\} =\bigl\{ (\tilde{A},\tilde{x}), (B,\tilde{y})\bigr\} . $$

(7.173)

Let

$$ u \in L_1 \hbox{ be such that } f(l_1-u) \le\rho_f(l_1,L_1)+2. $$

(7.174)

By (7.174), (7.173) and property (P2),

$$ \|u\| \le2a_m. $$

(7.175)

Relations (7.174), (7.173), (7.175), (7.165) and (7.128) imply that

$$\begin{aligned} \rho(u,L_2) =&\bigl|\rho(u,L_1)-\rho(u,L_2)\bigr| \le h_{2a_m}(L_1,L_2) \\ \le&h_{4a_m+4}(\tilde{A},B) \le2^{4a_m+5}\bar{\alpha}r. \end{aligned}$$

When combined with (7.132) and (7.134), this inequality implies that there is v∈X such that

$$ v \in L_2\quad \hbox{and}\quad \|u-v\| \in2^{4a_m+6} \bar{\alpha}r \le1. $$

(7.176)

Inequalities (7.175) and (7.176) imply that

$$ \|v\| \le1+2a_m. $$

(7.177)

By (7.177), (7.175), (7.173) and (7.163),

$$ \|l_1-u\|, \|l_2-v\| \le1+2a_m+m+1<3a_m. $$

(7.178)

It follows from (7.176), (7.139) and (7.173) that

$$ \bigl\| (l_1-u)-(l_2-v)\bigr\| \le\bar{\alpha}r+2^{4a_m+6} \bar{\alpha}r. $$

(7.179)

By (7.179), (7.178), (7.134) and the definition of k _m (see (7.131)),

$$\begin{aligned} \bigl|f(l_1-u)-f(l_2-v)\bigr| \le& k_m \bigl\| (l_1-u)-(l_2-v)\bigr\| \\ \le& k_m \bar{\alpha}r\bigl(1+2^{4a_m+6}\bigr) \le r \alpha2^{-9}. \end{aligned}$$

(7.180)

Inequalities (7.180) and (7.176) imply that

$$\rho_f(l_2,L_2) \le f(l_2-v) \le f(l_1-u)+2^{-9} \alpha r $$

and

$$ \rho_f(l_2,L_2) \le2^{-9}\alpha r+f(l_1-u). $$

(7.181)

Since (7.181) holds for any u satisfying (7.174), we conclude that

$$\rho_f(l_2,L_2) \le2^{-9} \alpha r+\rho_f(l_1,L_1). $$

This fact implies, in turn, that

$$ \bigl|\rho_f(l_1,L_1)- \rho_f(l_2,L_2)\bigr|=\bigl|\rho_f( \tilde{x},\tilde{A}) -\rho_f(\tilde{y},B)\bigr| \le2^{-9}\alpha r. $$

(7.182)

By property (P2), (7.169) and (7.140),

$$ \|\tilde{y}-z\| \le a_m \quad\hbox{and}\quad \|z\| \le2a_m. $$

(7.183)

It follows from (7.140), (7.183), (7.165) and (7.128) that

$$\begin{aligned} \rho(z,\tilde{A}) \le&\rho(z,B)+\bigl|\rho(z,B)-\rho(z,\tilde{A})\bigr| \\ =&\bigl|\rho(z,B)-\rho(z,\tilde{A})\bigr| \le h_{4a_m+4}(\tilde{A},B) \le 2^{4a_m+5}\bar{\alpha}r. \end{aligned}$$

Thus there exists $\tilde{z} \in X$ such that

$$ \tilde{z} \in\tilde{A} \quad\hbox{and}\quad \|z-\tilde{z}\| \le2^{4a_m+6}\bar{\alpha}r. $$

(7.184)

By (7.136), (7.183), (7.184), (7.134) and (7.132), we have

$$\begin{aligned} \|\tilde{x}-\tilde{z}\| \le&\|\tilde{x}\|+\|\tilde{z}\| \le m+\|z\|+ \|\tilde{z}-z\| \\ \le&m+2a_m+2^{4a_m+6}\bar{\alpha}r \le3a_m+1. \end{aligned}$$

When combined with (7.134), (7.184), (7.139), (7.140) and (7.182), this inequality implies that

$$\begin{aligned} f(\tilde{x}-\tilde{z}) \le& f(\tilde{y}-z)+\bigl|f(\tilde{x}-\tilde{z})-f(\tilde{y}-z)\bigr| \\ \le& f(\tilde{y}-z)+k_m\bigl\| \tilde{x}-\tilde{z}-(\tilde{y}-z)\bigr\| \le f(\tilde{y}-z) \\ \le& k_m\|\tilde{x}-\tilde{y}\|+k_m\|\tilde{z}-z\| \le f( \tilde{y}-z)+ k_m\bar{\alpha}r+k_m2^{4a_m+6} \bar{\alpha}r \\ \le&\rho_f(\tilde{y},B)+\alpha r+k_m\bar{\alpha}r \bigl(1+2^{4a_m+6}\bigr) \\ \le&\alpha r+k_m\bar{\alpha}r\bigl(1+2^{4a_m+6}\bigr)+ \rho_f(\tilde{x}, \tilde{A})+2^{-9}\alpha r \le\alpha r+ \alpha r+\rho_f(\tilde{x}, \tilde{A}). \end{aligned}$$

Thus we see that

$$ f(\tilde{x}-\tilde{z}) \le\rho_f(\tilde{x},\tilde{A})+2 \alpha r. $$

(7.185)

It follows from property (P1), (7.152), (7.185) and (7.184) that

$$\|\tilde{z}-\bar{x}\|\le(2n)^{-1}. $$

When combined with (7.184), (7.134) and (7.132), this inequality implies that

$$\|z-\bar{x}\| \le\|z-\tilde{z}\|+\|\tilde{z}-\bar{x}\|\le2^{4a_m+6}\bar{\alpha}r+ (2n)^{-1}\le n^{-1}. $$

Thus (7.142) is proved. Inequality (7.141) follows from (7.138), (7.151), (7.134) and (7.132). Thus we have shown that (7.138)–(7.140) imply (7.141) and (7.142). Lemma 7.20 is proved. □

7.13 Proofs of Theorems 7.16–7.18

We use the notations and the definitions from the previous section.

For each natural number n, denote by ${\mathcal{F}}_{n}$ the set of all (x,A)∈X×S(X) such that the following property holds:

(P3):

There exist y∈A and δ>0 such that for each $\tilde{x} \in X$ satisfying $\|\tilde{x}-x\|\le\delta$, each B∈S(X) satisfying h(A,B)≤δ, and each z∈B satisfying $f(\tilde{x}-z) \le\rho_{f}(\tilde{x},B)+\delta$, the inequality ∥z−y∥≤n ⁻¹ holds.

Set

$$ {\mathcal{F}}=\bigcap_{n=1}^{\infty} {\mathcal{F}}_n . $$

(7.186)

Lemma 7.21

If

$$ (x,A) \in{ \mathcal{F}}, $$

(7.187)

then the problem f(x−y)→min, y∈A, is strongly well posed.

Proof

Let $(x,A) \in{ \mathcal{F}}$ and let n be a natural number. Since $(x,A) \in{ \mathcal{F}} \subset{ \mathcal{F}}_{n}$, there exist x _n ∈A and δ _n >0 such that the following property holds:

(P4):

For each $\tilde{x} \in X$ satisfying $\|\tilde{x}-x\| \le \delta_{n}$, each B∈S(X) satisfying h(A,B)≤δ _n , and each z∈B satisfying $f(\tilde{x}-z) \le\rho_{f}(\tilde{x},B)+\delta_{n}$, the inequality ∥z−x _n ∥≤n ⁻¹ holds.

Suppose that

$$ \{z_i\}_{i=1}^{\infty} \subset A \quad\hbox{and}\quad \lim_{i \to\infty}f(x-z_i)= \rho_f(x,A). $$

(7.188)

Let n be any natural number. By (7.188) and property (P4), for all sufficiently large i we have

$$ f(x-z_i) \le\rho_f(x,A)+ \delta_n \quad\hbox{and}\quad \|z_i-x_n\|\le n^{-1}. $$

(7.189)

The second inequality of (7.189) implies that $\{z_{i}\}_{i=1}^{\infty}$ is a Cauchy sequence and there exists

$$ \bar{x}=\lim_{i \to\infty} z_i. $$

(7.190)

Limits (7.190) and (7.188) imply that

$$f(x-\bar{x}) =\rho_f(x,A). $$

Clearly, $\bar{x}$ is the unique solution of the problem f(x−z)→min, z∈A. Otherwise we would be able to construct a nonconvergent sequence $\{z_{i}\}_{i=1}^{\infty}$ satisfying (7.188). By (7.190) and (7.189),

$$ \|\bar{x}-x_n\|\le n^{-1},\quad n=1,2,\dots. $$

(7.191)

Let ε>0 be given. Choose a natural number

$$ n>8\varepsilon ^{-1}. $$

(7.192)

Assume that

$$\begin{gathered} \tilde{x} \in X, \quad\|\tilde{x}-x\| \le\delta_n,\quad\quad B \in S(X),\quad h(A,B) \le \delta_n, \\ z \in B, \quad\hbox{and} \quad f(\tilde{x}-z) \le\rho_f(\tilde{x},B)+ \delta_n. \end{gathered} $$

By Property (P4), ∥z−x _n ∥≤1/n. When combined with (7.192) and (7.191), this inequality implies that

$$\|z-\bar{x}\| \le\|z-x_n\|+\|x_n-\bar{x}\| \le(2n)^{-1}<\varepsilon . $$

Thus the problem f(x−z)→min, z∈A, is strongly well posed. Lemma 7.21 is proved. □

Proof of Theorem 7.16

For each integer n≥1, set

$$ \varOmega _n:=\bigl\{ A \in S(X): (\tilde{x},A) \in{ \mathcal{F}}_n\bigr\} $$

(7.193)

and let

$$ \varOmega :=\bigcap_{n=1}^{\infty} \varOmega _n. $$

(7.194)

By Lemma 7.21, (7.193) and (7.194), for each A∈Ω, the problem $f(\tilde{x}-z) \to\min$, z∈A, is strongly well posed. In order to prove the theorem, it is sufficient to show that for each natural number n, the set S(X)∖Ω _n is σ-porous with respect to $(h,\tilde{H})$. To this end, let n be any natural number.

Fix a natural number

$$ m_0>\|\tilde{x}\|. $$

(7.195)

For each integer m≥m ₀, define

$$ E_m:=\bigl\{ A \in S(X): A \cap\bigl\{ z \in X: \|z\| \le m \bigr\} \neq\emptyset \bigr\} . $$

(7.196)

Since

$$S(X) \setminus \varOmega _n=\bigcup_{m=m_0}^{\infty}(E_m \setminus \varOmega _n), $$

in order to prove the theorem, it is sufficient to show that for any natural number m≥m ₀, the set E _m ∖Ω _n is porous with respect to $(h,\tilde{H})$. Let m≥m ₀ be a natural number. Define

$$ \alpha_*=\bar{\alpha}(m+1,n)/2 $$

(7.197)

(see (7.132) and (7.134)). Let A∈S(X) and r∈(0,1]. There are two cases:

case (1), where

$$ A \cap\bigl\{ z \in X:\|z\| \le m+1\bigr\} =\emptyset $$

(7.198)

and case (2), where

$$ A \cap\bigl\{ z \in X: \|z\|\le m+1\bigr\} \neq\emptyset. $$

(7.199)

Consider the first case.

Let

$$ B \in S(X) \hbox{ be such that } h(A,B) \le2^{-m-2}. $$

(7.200)

We claim that B∉E _m . Assume the contrary. Then there is u∈X such that

$$ u \in B \quad\hbox{and}\quad \|u\| \le m. $$

(7.201)

By (7.201) and (7.128),

$$ \rho(u,A)\le\rho(u,B) +\bigl|\rho(u,B)-\rho(u,A)\bigr| \le h_m(A,B). $$

(7.202)

The definition of h _m (see (7.128)) and (7.200) imply that

$$\begin{gathered} h_m(A,B) \bigl(1+h_m(A,B)\bigr)^{-1}\le h(A,B)2^m \le2^{-2}, \\ h_m(A,B)\le h_m(A,B)2^{-2}+2^{-2} \end{gathered} $$

and

$$h_m(A,B) \le1/3. $$

When combined with (7.202), this implies that there is v∈A such that ∥u−v∥≤1/2. Together with (7.201) this inequality implies that ∥v∥≤m+1/2, a contradiction (see (7.198)). Therefore B∉E _m , as claimed. Thus we have shown that

$$ \bigl\{ B \in S(X): h(A,B) \le2^{-m-2}\bigr\} \cap E_m=\emptyset. $$

(7.203)

Now consider the second case. Then by Lemma 7.20, (7.195) and (7.199), there exists $\bar{x} \in X$ such that

$$\rho(\bar{x},A)\le r/8 $$

and such that for the set $\tilde{A}=A \cup\{\bar{x}\}$, the following property holds:

(P5):

if B∈S(X), $h(\tilde{A},B) \le\bar{\alpha}(m+1,n)r$, $\tilde{y} \in X$, $\|\tilde{y}-\tilde{x}\|\le\bar{\alpha}(m+1,n)r$, and z∈B satisfies

$$f(\tilde{y}-z) \le\rho_f(\tilde{y},B)+\bar{\alpha}(m+1,n), $$

then

$$\|z-\bar{x}\| \le n^{-1} \quad\hbox{and} \quad h(A,B) \le r. $$

Clearly,

$$\tilde{H}(A,\tilde{A}) \le r/8. $$

Property (P5), (7.193) and the definition of ${\mathcal{F}}_{n}$ (see (P3)) imply that

$$\bigl\{ B \in S(X): h(\tilde{A},B) \le\bar{\alpha}(m+1,n)r/2\bigr\} \subset \varOmega _n. $$

Thus in both cases we have

$$ \bigl\{ B \in S(X): h(\tilde{A},B) \le\alpha_*r/2\bigr\} \cap(E_m \setminus \varOmega _n) =\emptyset. $$

(7.204)

(Note that in the first case (7.204) is true with $\tilde{A}=A$.)

Therefore we have shown that the set E _m ∖Ω _n is porous with respect to $(h,\tilde{H})$. Theorem 7.16 is proved. □

Proof of Theorem 7.17

By Lemma 7.21, in order to prove the theorem, it is sufficient to show that for any natural number n, the set $(X \times S(X)) \setminus{ \mathcal{F}}_{n}$ is σ-porous in X×S(X) with respect to $(h,\tilde{H})$. To this end, let n be a natural number. For each natural number m, define

$$ E_m=\bigl\{ (x,A) \in X \times S(X): \|x\|\le m \hbox{ and } A \cap\bigl\{ z \in X: \|z\|\le m\bigr\} \neq\emptyset\bigr\} . $$

(7.205)

Since

$$\bigl(X \times S(X)\bigr) \setminus {\mathcal{F}}_n=\bigcup_{m=1}^{\infty} E_m \setminus{ \mathcal{F}}_n, $$

in order to prove the theorem it is sufficient to show that for each natural number m, the set $E_{m} \setminus{ \mathcal{F}}_{n}$ is porous in X×S(X) with respect to $(h, \tilde{H})$.

Let m be a natural number. Define α _∗ by (7.197). Assume that $(\tilde{x}\times A) \in X \times S(X)$ and r∈(0,1].

There are three cases:

case (1), where

$$\|\tilde{x}\|>m+1, $$

case (2), where

$$ \|\tilde{x}\| \le m+1 \quad\hbox{and}\quad \bigl\{ z \in A: \|z\| \le m+1\bigr\} = \emptyset, $$

(7.206)

and case (3), where

$$ \|\tilde{x}\| \le m+1 \quad\hbox{and} \quad\bigl\{ z \in A: \|z\| \le m+1\bigr\} \neq \emptyset. $$

(7.207)

In the first case,

$$ \bigl\{ (y,B) \in X \times S(X): d_1\bigl((\tilde{x},A),(y,B)\bigr) \le2^{-1}\bigr\} \cap E_m=\emptyset. $$

(7.208)

Next, consider the second case. In the proof of Theorem 7.16 we have shown that

$$\begin{aligned} &\hbox{if } B \in S(X) \hbox{ satisfies } h(A,B) \le2^{-m-2}, \hbox{ then} \\ &B \cap\bigl\{ z \in X: \|z\| \le m\bigr\} =\emptyset \end{aligned}$$

and

$$ \bigl\{ (y,B) \in X \times S(X): d_1\bigl((y,B),( \tilde{x},A)\bigr) \le2^{-m-2}\bigr\} \cap E_m= \emptyset. $$

(7.209)

Finally, consider the third case. Then by Lemma 7.20, there exists $\bar{x} \in X$ such that $\rho(\bar{x},A) \le r/8$ and such that for the set $\tilde{A}=A\cup\{\bar{x}\}$, property (P5) holds. Clearly,

$$d_2\bigl((\tilde{x},A),(\tilde{x},\tilde{A})\bigr)=\tilde{H}(A,\tilde{A}) \le r/8. $$

Property (P5) implies that

$$\bigl\{ (\tilde{y},B) \in X \times S(X): d_1\bigl((\tilde{y},B),(\tilde{x},\tilde{A})\bigr) \le\bar{\alpha}(m+1,n)r/2\bigr\} \subset{ \mathcal{F}}_n. $$

Hence in all three cases we have

$$ \bigl\{ (\tilde{y},B) \in X \times S(X): d_1\bigl(( \tilde{y},B),(\tilde{x}, \tilde{A})\bigr) \le \alpha_*r\bigr\} \cap(E_m \setminus{ \mathcal{F}}_n)=\emptyset. $$

(7.210)

Note that in the first and second cases, (7.210) is true with $A=\tilde{A}$. Therefore we have shown that the set $E_{m}\setminus{ \mathcal{F}}_{n}$ is porous with respect to (d ₁,d ₂). Theorem 7.17 is proved. □

Proof of Theorem 7.18

Let $\{x_{i}\}_{i=1}^{\infty}$ be a countable dense subset of X ₀. By countable dense subset of X ₀. By Theorem 7.16, for each ${\mathcal{F}}_{i} \subset S(X)$ such that $S(X) \setminus{ \mathcal{F}}_{i}$ is σ-porous in S(X) with respect to $(h,\tilde{H})$ and such that for each A∈S(X), the problem f(x _i −z)→min, z∈X, is strongly well posed. Set

$$ {\mathcal{F}}:=\bigcap_{i=1}^{\infty} {\mathcal{F}}_i. $$

(7.211)

Clearly, $S(X) \setminus{ \mathcal{F}}$ is a σ-porous subset of S(X) with respect to $(h,\tilde{H})$.

Let $A \in{ \mathcal{F}}$. Assume that n and i are natural numbers. Since the problem f(x _i −z)→min, z∈A, is strongly well posed, there exists a number δ _in >0 and a unique $\bar{x}_{i} \in A$ such that

$$ f(x_i-\bar{x}_i)=\rho_f(x_i,A) $$

(7.212)

and the following property holds:

(P6):

if y∈X satisfies ∥y−x _i ∥≤δ _in , B∈S(X) satisfies h(A,B)≤δ _in , and z∈B satisfies

$$ f(y-z) \le\rho_f(y,B) +\delta_{in}, $$

(7.213)

then $\|z-\bar{x}_{i}\| \le(2n)^{-1}$.

Define

$$\begin{aligned} F=\bigcap_{q=1}^{\infty} \bigcup\bigl\{ \bigl\{ z \in X: \|z-x_i\|<\delta_{in}\bigr\} : i=1,2,\dots, n=q,q+1,\dots\bigr\} \cap X_0. \end{aligned}$$

(7.214)

Clearly, F is a countable intersection of open everywhere dense subsets of X ₀. Let

$$ \tilde{x} \in F. $$

(7.215)

For each natural number q, there exist natural numbers n _q ≥q and i _q such that

$$ \|\tilde{x}-x_{iq}\| <\delta_{i_q n_q}. $$

(7.216)

Assume that

$$ \{y_k\}_{k=1}^{\infty} \subset A \quad\hbox{and}\quad \lim_{k \to\infty}f(\tilde{x}-y_k)= \rho_f(\tilde{x},A). $$

(7.217)

Let q be a natural number. Then for all sufficiently large natural numbers k,

$$f(\tilde{x}-y_k) \le\rho_f(\tilde{x},A)+ \delta_{i_qn_q}, $$

and by property (P6) and (7.216),

$$ \|y_k-\bar{x}_{i_q}\| \le(2n_q)^{-1} \le(2q)^{-1}. $$

(7.218)

This implies that $\{y_{k}\}_{k=1}^{\infty}$ is a Cauchy sequence and there exists $\bar{x} =\lim_{k \to\infty}y_{k}$. By (7.217), $f(\tilde{x}-\bar{x})= \rho_{f}(\tilde{x},A)$. Clearly, $\bar{x}$ is the unique minimizer for the problem $f(\tilde{x}-z) \to\min$, z∈A. Otherwise, we would be able to construct a nonconvergent sequence $\{y_{k}\}_{k=1}^{\infty}$. By (7.218),

$$ \|\bar{x}-x_{i_q}\| \le(2q)^{-1},\quad q=1,2,\dots. $$

(7.219)

Let ε>0 be given. Choose a natural number

$$q>8\varepsilon ^{-1}. $$

Set

$$ \delta=\delta_{i_qn_q}-\|\tilde{x}-x_{i_q}\|. $$

(7.220)

By (7.216), δ>0. Assume that

$$ y \in X,\quad \|y-\tilde{x}\| \le\delta,\quad\quad B \in S(X),\quad h(A,B) \le\delta, $$

(7.221)

and

$$z \in B, \quad f(y-z) \le\rho_f(y,B)+\delta. $$

By (7.220) and (7.221),

$$ \|y-x_{i_q}\|\le\|y-\tilde{x}\|+\|\tilde{x}-x_{i_q}\| \le \delta_{i_qn_q}. $$

(7.222)

By (7.222), (7.220) and property (P6), $\|z-\bar{x}_{i_{q}}\| \le(2q)^{-1}$. When combined with (7.219), this inequality implies that $\|z-\bar{x}\| \le q^{-1}<\varepsilon$. This completes the proof of Theorem 7.18. □